## The equation of the straight line(s) that touches both x^2 + y^2 = 2a^2 and y^2 = 8ax is /are

Q: The equation of the straight line(s) that touches both x2 + y2 = 2a2 and y2 = 8ax is /are
(A) y = x + 2a

(B) y = – x – 2a

(C) y = – x + 2a

(D) y = x – 2a

Sol. The equation of any tangent to the circle x2 + y2 = 2a2 is

xcosθ + ysinθ = √2a . . . . . (1)

The equation of any tangent to the parabola y2 = 8ax is

y = mx + 2a/m . . . . . (2)

Since (1) and (2) are identical

$\large \frac{cos\theta}{-m} = \frac{sin\theta}{1} = \frac{m}{\sqrt{2}}$

$\large cos\theta = \frac{-m^2}{\sqrt{2}}$ and $\large sin\theta = \frac{m}{\sqrt{2}}$

Squaring and adding, m4 + m2 – 2 = 0

⇒ m2 = 1 ⇒ m = ± 1

Substituting in (2) the equation of the required tangent is y = ± (x + 2a)

Hence (A) and (B) are the correct answers.

## If two chords drawn from the point (4 , 4) to the parabola x^2 = 4y are divided by line y = m x in the ratio 1 : 2 , then

Q: If two chords drawn from the point (4 , 4) to the parabola x^2 = 4y are divided by line y = m x in the ratio 1 : 2 , then

(A) m < – √3

(B) m < –√3 – 1

(C) m > √3

(D) m > √3– 1

Sol. Point (4, 4) lies on the parabola. Let the point of intersection of the line y = m x with the chords be (a , ma), then

$\large \alpha = \frac{4+2x_1}{3}$

$\large x_1 = \frac{3\alpha – 4}{2}$

and , $\large m \alpha = \frac{4 + 2 y_1}{3}$

$\large y_1 = \frac{3m \alpha – 4}{2}$

as (x1, y1) lies on the curve

$\large (\frac{3\alpha – 4}{2})^2 = 4 (\frac{3m\alpha – 4}{2})$

⇒ 9α2 + 16 – 24a = 8(3mα – 4)

Þ 9α2 – 24α(1 + m) + 48 = 0

Þ 3α2 – 8α(1 + m) + 16 = 0

two distinct chords are obtained

D > 0.

(8(1 + m))2 – 4 . 3 . 16 > 0

⇒ (1 + m)2 – 3 > 0

1 + m > √3 or 1 + m < –√3

⇒ m > √3– 1 or m < –( √3 + 1).

Hence (B), (C) and (D) are the correct answers.

## If normal chord at the point (t^2, 2t) of the parabola y^2 = 4x subtends a right angle at the vertex then the value of t is/are

Q: If normal chord at the point (t2 , 2t) of the parabola y2 = 4x subtends a right angle at the vertex then the value of t is/are

(A) √3

(B) √2

(C) -√3

(D) -√2

Sol. Normal at t t1 = -t -2/t …(1)

AB ⊥ BC

⇒ $\large \frac{2t – 0}{t^2 – 0} . \frac{2t_1 – 0}{t_1^2 – 0} = -1$

-4tt1 = (tt1)2

t1 t = -4

⇒ t2 + 2 = 4

⇒ t = ± √2

Hence (B) and (D) are the correct answers.

## Let S be the set of all possible values of parameter ‘a’ for which the points of intersection of the parabolas y^2 = 3ax and

Q: Let S be the set of all possible values of parameter ‘a’ for which the points of intersection of the parabolas y2 = 3ax and  $y = \frac{1}{2}(x^2 + ax + 5)$ y = are concyclic . Then S contains the interval(s)

(A) (– ∞ , – 2)

(B) (– 2, 0)

(C) (0, 2)

(D) (2, ∞)

Ans: (A), (D)
Sol: Family of curves passing the points of intersection of two parabolas is

y2 – 3ax + λ(x2 + ax + 5 – 2y) = 0

it will represents a circle if  λ = 1

x2 + y2 – 2ax – 2y + 5 = 0

it represents a real circle if a2 + 1 – 5 > 0

⇒ a2 > 4

⇒ a ∈ (-∞, -2) ∪ (2, ∞).

Hence (A) and (D) are the correct answers.

## The equation 2y^2 + 3y – 4x – 3 = 0 represents a parabola for which

Q: The equation 2y2 + 3y – 4x – 3 = 0 represents a parabola for which

(A) length of latus rectum is 4

(B) equation of the axis is 4y + 3 = 0

(C) equation of directrix is 2x + 1 = 0

(D) equation of tangent at vertex is x = -33/32

Ans: (B) & (D)
Sol: The given equation can be re-written as $\large (y+\frac{3}{4})^2 = 2(x+\frac{33}{32})$

which is of the form Y2 = 4aX.

Hence the vertex is $(-\frac{33}{32} , -\frac{3}{4})$

The axis is y + 3/4 = 0

⇒ y = -3/4

The directrix is x + a = 0

⇒ x +33/32 +1/2 = 0

⇒ x = -49/32

The tangent at the vertex is x + 33/32 = 0

⇒ x = -33/32

Length of the latus rectum = 4a = 2.

Hence (B) and (D) are the correct answers.