RS Aggarwal Solutions for Class 11 Chapter 22 Parabola

Question 1. Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: \( y^2 = 12x \)
Answer: Start with the equation \( y^2 = 12x \). Match this to the standard form \( y^2 = 4ax \). This gives \( 4a = 12 \), so \( a = 3 \).

The focus is at \( F(a,0) = F(3,0) \).

The vertex sits at \( A(0,0) \).

For the directrix, use \( x + a = 0 \), which becomes \( x + 3 = 0 \), so \( x = -3 \).

The axis is the x-axis (the line \( y = 0 \)).

The length of the latus rectum is \( 4a = 4(3) = 12 \).
In simple words: When you have an equation like \( y^2 = 12x \), you can find special points and lines using a simple formula. The focus is the "special point" the parabola wraps around, the directrix is a "special line" it stays away from, and the latus rectum measures how wide the parabola opens.

Exam Tip: Always identify the value of \( a \) first by comparing with the standard form - this one number gives you everything else you need. Double-check your directrix equation by confirming the focus and vertex are equidistant from it.

Question 2. Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: \( y^2 = 10x \)
Answer: Start with the equation \( y^2 = 10x \). Match this to the standard form \( y^2 = 4ax \). This gives \( 4a = 10 \), so \( a = 2.5 \).

The focus is at \( F(a,0) = F(2.5,0) \).

The vertex sits at \( A(0,0) \).

For the directrix, use \( x + a = 0 \), which becomes \( x + 2.5 = 0 \), so \( x = -2.5 \).

The axis is the x-axis (the line \( y = 0 \)).

The length of the latus rectum is \( 4a = 4(2.5) = 10 \).
In simple words: Use the same process as before: extract \( a \) from the equation, then plug it into the formulas for the focus, directrix, and latus rectum length.

Exam Tip: When \( a \) is a decimal or fraction, keep it in that form throughout - do not round until the final answer if requested.

Question 3. Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: \( 3y^2 = 8x \)
Answer: Start with the equation \( 3y^2 = 8x \). Rearrange to isolate \( y^2 \): \( y^2 = \frac{8}{3}x \). Match this to the standard form \( y^2 = 4ax \). This gives \( 4a = \frac{8}{3} \), so \( a = \frac{2}{3} \).

The focus is at \( F(a,0) = F\left(\frac{2}{3},0\right) \).

The vertex sits at \( A(0,0) \).

For the directrix, use \( x + a = 0 \), which becomes \( x + \frac{2}{3} = 0 \), so \( x = -\frac{2}{3} \).

The axis is the x-axis (the line \( y = 0 \)).

The length of the latus rectum is \( 4a = \frac{8}{3} \).
In simple words: When the parabola equation has a coefficient on \( y^2 \), divide both sides to get \( y^2 \) by itself first. Then proceed exactly as before.

Exam Tip: Never forget the rearrangement step when there is a coefficient - this is a common error that flips the whole problem.

Question 4. Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: \( y^2 = -8x \)
Answer: Start with the equation \( y^2 = -8x \). Match this to the standard form \( y^2 = -4ax \). This gives \( 4a = 8 \), so \( a = 2 \).

The focus is at \( F(-a,0) = F(-2,0) \).

The vertex sits at \( A(0,0) \).

For the directrix, use \( x - a = 0 \), which becomes \( x - 2 = 0 \), so \( x = 2 \).

The axis is the x-axis (the line \( y = 0 \)).

The length of the latus rectum is \( 4a = 8 \).
In simple words: When the equation has a negative sign (like \( y^2 = -8x \)), the parabola opens to the left instead of to the right. The focus sits on the negative x-axis, and the directrix is on the positive x-axis.

Exam Tip: The negative sign in the equation tells you the parabola direction - pay close attention to signs when setting up your focus and directrix coordinates.

Question 5. Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: \( y^2 = -6x \)
Answer: Start with the equation \( y^2 = -6x \). Match this to the standard form \( y^2 = -4ax \). This gives \( 4a = 6 \), so \( a = \frac{3}{2} \).

The focus is at \( F(-a,0) = F\left(-\frac{3}{2},0\right) \).

The vertex sits at \( A(0,0) \).

For the directrix, use \( x - a = 0 \), which becomes \( x - \frac{3}{2} = 0 \), so \( x = \frac{3}{2} \).

The axis is the x-axis (the line \( y = 0 \)).

The length of the latus rectum is \( 4a = 6 \).
In simple words: This parabola also opens to the left. Extract \( a \) from the equation and use the formulas for the left-opening case.

Exam Tip: Compare your result against the graph to confirm the focus is indeed on the correct side and the directrix is at the right location.

Question 6. Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: \( 5y^2 = -16x \)
Answer: Start with the equation \( 5y^2 = -16x \). Rearrange to isolate \( y^2 \): \( y^2 = -\frac{16}{5}x \). Match this to the standard form \( y^2 = -4ax \). This gives \( 4a = \frac{16}{5} \), so \( a = \frac{4}{5} \).

The focus is at \( F(-a,0) = F\left(-\frac{4}{5},0\right) \).

The vertex sits at \( A(0,0) \).

For the directrix, use \( x - a = 0 \), which becomes \( x - \frac{4}{5} = 0 \), so \( x = \frac{4}{5} \).

The axis is the x-axis (the line \( y = 0 \)).

The length of the latus rectum is \( 4a = \frac{16}{5} \).
In simple words: First rearrange the equation by dividing both sides by 5 to get \( y^2 \) alone. Then identify that this is a left-opening parabola and apply the formulas.

Exam Tip: When a coefficient appears on \( y^2 \) in a negative parabola, handle the division carefully to avoid sign errors.

Question 7. Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: \( x^2 = 16y \)
Answer: Start with the equation \( x^2 = 16y \). Match this to the standard form \( x^2 = 4ay \). This gives \( 4a = 16 \), so \( a = 4 \).

The focus is at \( F(0,a) = F(0,4) \).

The vertex sits at \( A(0,0) \).

For the directrix, use \( y + a = 0 \), which becomes \( y + 4 = 0 \), so \( y = -4 \).

The axis is the y-axis (the line \( x = 0 \)).

The length of the latus rectum is \( 4a = 16 \).
In simple words: When \( x^2 \) is by itself (not \( y^2 \)), the parabola opens either up or down instead of left or right. This one opens upward. The focus is above the vertex, and the directrix is below it.

Exam Tip: Notice the shift in roles: now the y-coordinate of the focus changes (not the x-coordinate), and the directrix is a horizontal line (not vertical).

Question 8. Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: \( x^2 = 10y \)
Answer: Start with the equation \( x^2 = 10y \). Match this to the standard form \( x^2 = 4ay \). This gives \( 4a = 10 \), so \( a = 2.5 \).

The focus is at \( F(0,a) = F(0,2.5) \).

The vertex sits at \( A(0,0) \).

For the directrix, use \( y + a = 0 \), which becomes \( y + 2.5 = 0 \), so \( y = -2.5 \).

The axis is the y-axis (the line \( x = 0 \)).

The length of the latus rectum is \( 4a = 10 \).
In simple words: Apply the same upward-opening formulas as before, but with \( a = 2.5 \) this time.

Exam Tip: Verify your answer by sketching - the focus should sit directly above the vertex on the axis of symmetry.

Question 9. Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: \( 3x^2 = 8y \)
Answer: Start with the equation \( 3x^2 = 8y \). Rearrange to isolate \( x^2 \): \( x^2 = \frac{8}{3}y \). Match this to the standard form \( x^2 = 4ay \). This gives \( 4a = \frac{8}{3} \), so \( a = \frac{2}{3} \).

The focus is at \( F(0,a) = F\left(0,\frac{2}{3}\right) \).

The vertex sits at \( A(0,0) \).

For the directrix, use \( y + a = 0 \), which becomes \( y + \frac{2}{3} = 0 \), so \( y = -\frac{2}{3} \).

The axis is the y-axis (the line \( x = 0 \)).

The length of the latus rectum is \( 4a = \frac{8}{3} \).
In simple words: Divide the equation by 3 first to get \( x^2 \) by itself. Then use the upward-opening formulas with the new value of \( a \).

Exam Tip: Always simplify coefficient ratios carefully when dividing - a single arithmetic slip changes every answer that follows.

Question 10. Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: \( x^2 = -8y \)
Answer: Start with the equation \( x^2 = -8y \). Match this to the standard form \( x^2 = -4ay \). This gives \( 4a = 8 \), so \( a = 2 \).

The focus is at \( F(0,-a) = F(0,-2) \).

The vertex sits at \( A(0,0) \).

For the directrix, use \( y - a = 0 \), which becomes \( y - 2 = 0 \), so \( y = 2 \).

The axis is the y-axis (the line \( x = 0 \)).

The length of the latus rectum is \( 4a = 8 \).
In simple words: When the equation has a negative sign, the parabola opens downward instead of upward. The focus is below the vertex, and the directrix is above it.

Exam Tip: The negative coefficient always reverses the opening direction - down instead of up for vertical parabolas, left instead of right for horizontal ones.

Question 11. Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: \( x^2 = -18y \)
Answer: Start with the equation \( x^2 = -18y \). Match this to the standard form \( x^2 = -4ay \). This gives \( 4a = 18 \), so \( a = \frac{9}{2} \).

The focus is at \( F(0,-a) = F\left(0,-\frac{9}{2}\right) \).

The vertex sits at \( A(0,0) \).

For the directrix, use \( y - a = 0 \), which becomes \( y - \frac{9}{2} = 0 \), so \( y = \frac{9}{2} \).

The axis is the y-axis (the line \( x = 0 \)).

The length of the latus rectum is \( 4a = 18 \).
In simple words: This downward-opening parabola has a larger value of \( a \) compared to the previous one, which means the focus is farther from the vertex and the parabola opens more gradually.

Exam Tip: A larger value of \( a \) means a wider, more gradual opening - a smaller \( a \) produces a narrower, sharper curve.

Question 4. Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: 3x² = -16y
Answer: Starting with the equation 3x² = -16y, we can rewrite it as \( x^2 = -\frac{16}{3}y \).

Comparing this with the standard form \( x^2 = 4ay \), we get \( 4a = -\frac{16}{3} \), so \( a = -\frac{4}{3} \).

For a parabola of the form \( x^2 = 4ay \):
- Vertex: \( A(0, 0) \)
- Focus: \( F(0, a) = F(0, -\frac{4}{3}) \)
- Equation of directrix: \( y - a = 0 \), which gives \( y = \frac{4}{3} \)
- Axis of symmetry: the y-axis, or \( x = 0 \)
- Length of latus rectum: \( |4a| = \frac{16}{3} \)

Exam Tip: Always convert the given equation to standard form first. The sign and position of the parameter \( a \) determine whether the parabola opens upward or downward, and this directly affects the focus and directrix positions.

Question 5. Find the equation of the parabola with vertex at the origin and focus at F(-2, 0).
Answer: The vertex is at A(0, 0) and the focus is given as F(-2, 0). Since the focus lies on the x-axis, the parabola has a horizontal axis of symmetry.

For a parabola with vertex at the origin and focus at F(-a, 0), the standard form is \( y^2 = -4ax \).

Here, comparing F(-2, 0) with F(-a, 0), we get \( a = 2 \).

Therefore, the equation of the parabola is \( y^2 = -8x \).

Exam Tip: When the focus is to the left of the vertex (negative x-coordinate), the parabola opens leftward and the equation includes a negative sign before the linear term.

Question 6. Find the equation of the parabola with focus F(4, 0) and directrix x = -4.
Answer: The directrix is given as x = -4, which can be written as \( x + 4 = 0 \). This is of the form \( x + a = 0 \), giving us \( a = 4 \).

The focus is F(4, 0), which matches the form F(a, 0).

For a parabola with directrix \( x + a = 0 \) and focus at (a, 0), the standard equation is \( y^2 = 4ax \).

Substituting \( a = 4 \), we get \( y^2 = 16x \).

Exam Tip: Notice that the vertex lies exactly midway between the focus and directrix. Verify this quickly before finalizing your answer to avoid errors.

Question 7. Find the equation of the parabola with focus F(0, -3) and directrix y = 3.
Answer: The directrix is given as y = 3, which can be rewritten as \( y - 3 = 0 \). This is of the form \( y - a = 0 \), so \( a = 3 \).

The focus is F(0, -3), which matches the form F(0, -a).

For a parabola with directrix \( y - a = 0 \) and focus at (0, -a), the standard equation is \( x^2 = -4ay \).

Substituting \( a = 3 \), we get \( x^2 = -12y \).

Exam Tip: The parabola opens away from the directrix. Here, since the directrix is above the focus, the parabola opens downward, reflected in the negative coefficient of y.

Question 8. Find the equation of the parabola with vertex at the origin and focus F(0, 5).
Answer: The vertex is at A(0, 0) and the focus is at F(0, 5). Since the focus lies on the positive y-axis, the parabola has a vertical axis of symmetry and opens upward.

For a parabola with vertex at the origin and focus at F(0, a), the standard form is \( x^2 = 4ay \).

Here, comparing F(0, 5) with F(0, a), we get \( a = 5 \).

Therefore, the equation of the parabola is \( x^2 = 20y \).

Exam Tip: When the focus lies on the positive y-axis, the parabola opens upward. The parameter a represents the distance from the vertex to the focus, which is always positive in this standard form.

Question 9. Find the equation of the parabola with vertex at the origin, passing through the point P(5, 2) and symmetric with respect to the y-axis.
Answer: A parabola with vertex at the origin and symmetric about the y-axis has the standard form \( x^2 = 4ay \).

Since the parabola passes through point P(5, 2), we substitute these coordinates into the equation:
\( 5^2 = 4a(2) \)
\( 25 = 8a \)
\( a = \frac{25}{8} \)

Therefore, the equation becomes \( x^2 = 4 \cdot \frac{25}{8} \cdot y = \frac{25}{2}y \), or equivalently \( 2x^2 = 25y \).

Exam Tip: Always use the given point to find the unknown parameter. Substitute carefully and simplify the final equation to the cleanest form possible.

Question 10. Find the equation of the parabola, which is symmetric about the y-axis and passes through the point P(2, -3).
Answer: A parabola symmetric about the y-axis with vertex at the origin takes the form \( x^2 = 4ay \).

Substituting the point P(2, -3) into this equation:
\( 2^2 = 4a(-3) \)
\( 4 = -12a \)
\( a = -\frac{1}{3} \)

Therefore, the equation is \( x^2 = 4 \cdot (-\frac{1}{3}) \cdot y = -\frac{4}{3}y \), which simplifies to \( 3x^2 = -4y \).

Exam Tip: When the y-coordinate of the point is negative and a turns out negative, the parabola opens downward. Double-check the sign of a against the geometry of the problem for confirmation.