# Class 11 RD Sharma Solutions – Chapter 25 Parabola – Exercise 25.2

### Question 1. Find the axis of symmetry of the parabola y^{2} = x.

**Solution:**

We are given,

=> y

^{2}= xWe know this parabola lies on the y-axis.

Therefore, its vertex lies on the y-axis.

Now we know, parabola always has one axis which maintains its symmetry.

In this case, the axis is an x-axis, which divides the parabola into two equal parts.

Hence the required axis is x-axis.

### Question 2. Find the distance between the vertex and the focus of the parabola y^{2} + 6y + 2x + 5 = 0.

**Solution:**

Given that,

y

^{2}+ 6y + 2x + 5 = 0(y + 3)

^{2}+ 2x – 4 = 0(y + 3)

^{2}= -2 (x – 2)Let us assume Y = y + 3 and X = x – 2.

From the equation, we get,

Y

^{2}= – 2XOn putting 4a = 2, we get

=> a = 1/2

Focus = (X = -1/2, Y = 0) = (x = 3/2, y = -3)

Vertex = (X = 0, Y = 0) = (x = 2, y = -3)

Thus, we get

Focus = (3/2, -3)

Vertex = (2, -3)

The distance between the vertex and the focus is,

D =

=

= 1/2 units

Therefore, the required distance is 1/2 units.

### Question 3. Find the equation of the directrix of the parabola x^{2} − 4x − 8y + 12 = 0.

**Solution:**

Given that,

x

^{2 }− 4x − 8y + 12 = 0(x – 2)

^{2}– 4 – 8y + 12 = 0(x – 2)

^{2}= 8 (y – 1)Let Y = y − 1 and X = x – 2.

From the above equation, we get,

X

^{2}= 8YOn comparing with x

^{2}= 4ay, we get=> a = 2

Directrix = Y = −a

=> y − 1 = −a

=> y = −a + 1

=> y = −2 + 1

=> y = −1

Therefore, the equation of the directrix is y = -1.

### Question 4. Write the equation of the parabola with focus (0, 0) and directrix x + y − 4 = 0.

**Solution:**

Let us assume P (x, y) be any point on the parabola whose focus is S (0, 0).

And the equation of directrix is x + y= 4.

First we draw PM perpendicular to x + y = 4.

Then, we have,

SP = PM

SP

^{2}= PM^{2}(x – 0)

^{2}+ (y – 0)^{2}=x

^{2}+ y^{2}=2x

^{2}+ 2y^{2 }= x^{2}+ y^{2}+ 16 + 2xy – 8y – 8x=> x

^{2}+ y^{2}– 2xy + 8x + 8y – 16 = 0

Hence, the equation of the parabola is x^{2}+ y^{2}– 2xy + 8x + 8y – 16 = 0

### Question 5. Find the length of the chord of the parabola y^{2} = 4ax which passes through the vertex and is inclined to the axis at π/4.

**Solution:**

Let us considered OP be the chord and the coordinates of P be (x

_{1}, y_{1}).We have,

=> OP

^{2}= x_{1}^{2}+ y_{1}^{2}And, tan π/4 = y

_{1}/x_{1}=> 1 = y

_{1}/x_{1}=> x

_{1}= y_{1}and (x

_{1}, y_{1}) lies on the parabola.∴ y

_{1}^{2}= 4ax_{1}So, we get

=> x

_{1}^{2}= 4a x_{1 }=> x

_{1}= 4aTherefore, we have

OP

^{2}= (4a)^{2}+ (4a)^{2}OP

^{2}= 32a^{2}OP = 4√2 a

Therefore, the length of the chord is 4√2 a.

### Question 6. If b and c are the lengths of the segments of any focal chord of the parabola y^{2} = 4ax, then write the length of its latus-rectum.

**Solution:**

Let us assume S (a, 0) be the focus of the given parabola and the end points of the focal chord be P(at

^{2}, 2at) and Q (a/t^{2}}, -2a/t).Now, SP and SQ are segments of the focal chord with lengths b and c.

∴ SP = b, SQ = c

Also, SP =

= a (1 + t

^{2})And, SQ =

= a (1 + 1/t

^{2})Now, we get,

=> a =

Hencec, the length of the latus rectum is 4a =

### Question 7. PSQ is a focal chord of the parabola y^{2} = 8x. If SP = 6, then find SQ.

**Solution:**

According to the question, it is given that PSQ is a focal chord of the parabola y

^{2}= 8x and SP = 6So,

the coordinates of the focal chord are P (at^{2}, 2at) and Q (a/t^{2}, – 2a/t).On comparing y

^{2}= 8x with y^{2}= 4ax, we get,=> 4a = 8

=> a = 2

Therefore, the coordinates of the focus S is (2, 0).

SP = 6

t

^{4}+ 2t^{2}– 8 = 0=> t

^{2}= 2Thus, we get,

SQ =

=

=

= √5

Hence, SQ is √5.

### Question 8. Write the coordinates of the vertex of the parabola whose focus is at (−2, 1) and directrix is the line x + y − 3 = 0.

**Solution:**

Given that, the focus S is at (−2, 1) and the directrix is the line x + y − 3 = 0 and the slope of the line perpendicular to x + y − 3 = 0 is 1.

Now, the axis of the parabola is perpendicular to the directrix and passes through the focus.

So, the equation of the axis of the parabola becomes,

=> y – 1 = 1 (x + 2)

Now, the intersection point of the directrix and axis is the intersection point of parabola and x + y − 3 = 0.

Let us assume that the intersection point be K.

So, the coordinates of K are (0, 3).

Let us assume that the point (h, k) be the coordinates of the vertex, which is the mid-point of the line segment joining K and the focus.

=> h = (0 – 2)/2, k = (3 + 1)/2

=> h = -1, k = 2

Hence, the coordinates of the vertex are (−1, 2).

### Question 9. If the coordinates of the vertex and focus of a parabola are (−1, 1) and (2, 3) respectively, then write the equation of its directrix.

**Solution:**

Given that, the vertex and the focus of a parabola are (−1, 1) and (2, 3).

So, the slope of the axis of the parabola = = 2/3

Slope of the directrix = -3/2

Let us assume the directrix intersect the axis at point K (r, s).

So, (r + 2)/2 = -1, (s + 3)/2 = 1

=> r = -4, s = -1

Now, the required equation of the directrix is,

y + 1 = (-3/2) (x + 4)

=> 3x + 2y + 14 = 0

### Question 10. If the parabola y^{2} = 4ax passes through the point (3, 2), then find the length of its latus rectum.

**Solution:**

Given that, y

^{2}= 4ax and the parabola is passing through the point (3, 2)So, this will satisfy the equation of the parabola.

22 = 4(a)(3)

=> a = 1/3

The length of the latus ractum = 4a

= 4 (1/3)

= 4/3

Hence, the latus ractum is 4/3

### Question 11. Write the equation of the parabola whose vertex is at (−3, 0) and the directrix is x + 5 = 0.

**Solution:**

Given that the vertex(h, k) is (-3, 0) and the directrix is x + 5 = 0

So, the general equation of the parabola = (y − k)

^{2}= 4a (x − h).Now, the directrix is

x = h − a

As x + 5 = 0, we get

=> −5 = −3 − a

=> a = 2

So, the equation of the parabola is

(y − 0)

^{2}= 4 (2) (x + 3)=> y

^{2 }= 8 (x + 3)

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