Let the tangent and normal at the point ( 3 3 , 1 ) on the ellipse x 2 36 + y 2 4 = 1 meet the y -axis at the points A and B respectively. Let the circle C be drawn taking AB as a diameter and the line x = 2 5 intersect C at the points P and Q. If

Question

Let the tangent and normal at the point $(3 \sqrt{3}, 1)$ on the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{4} = 1$ meet the $y$ -axis at the points A and B respectively. Let the circle C be drawn taking AB as a diameter and the line $x = 2 \sqrt{5}$ intersect C at the points P and Q. If the tangents at the points P and Q on the circle intersect at the point $(α, β)$ , then $α^{2} - β^{2}$ is equal to [2023]

Smart Achievers · Accepted Answer

(3)

We have, $\frac{x^{2}}{36} + \frac{y^{2}}{4} = 1$ ...(i)

$∴$ The equation of tangent at point $(3 \sqrt{3}, 1)$ on the given ellipse is given by

$y - 1 = - 3 \sqrt{3} (\frac{4}{36}) (x - 3 \sqrt{3})$ $[∵ y - y_{1} = \frac{- x_{1}}{y_{1}} (\frac{b^{2}}{a^{2}}) (x - x_{1})]$

$\Rightarrow y - 1 = \frac{- x}{\sqrt{3}} + 3$ ...(i)

$\Rightarrow \sqrt{3} y - \sqrt{3} = - x + 3 \sqrt{3} \Rightarrow x + \sqrt{3} y = 4 \sqrt{3}$ ...(ii)

The equation of normal at point $(3 \sqrt{3}, 1)$ on the given ellipse

$y - 1 = \frac{1}{3 \sqrt{3}} (\frac{36}{4}) (x - 3 \sqrt{3})$ $[∵ y - y_{1} = (\frac{y_{1}}{x_{1}}) (\frac{a^{2}}{b^{2}}) (x - x_{1})]$

$\Rightarrow y - 1 = \sqrt{3} (x - 3 \sqrt{3}) \Rightarrow y - 1 = \sqrt{3} x - 9 \Rightarrow \sqrt{3} x - y = 8$

Tangent meets $y$ –axis at $y$ = 4 and normal meets $y$ –axis at $y$ = - 8.

$∴$ $A \equiv (0, 4)$ and $B \equiv (0, - 8)$

$∴$ $Equation of circle with diameter end points A and B is x^{2} + (y - 4) (y + 8) = 0$

$[∵ Equation of circle with diameter end points (x_{1}, y_{1}) is (x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0]$

$\Rightarrow x^{2} + y^{2} + 4 y - 32 = 0$

$∴$ $Centre of circle = (0, - 2)$

Equation of line passing through centre of circle, $y = - 2$ .

Now, the equation of chord of contact PQ of two tangents drawn from the point $(α, β)$ is given by

$α x + β y + 2 (y + β) - 32 = 0$

Since $(2 \sqrt{5}, 0)$ lie on the chord of contact PQ

$\Rightarrow α (2 \sqrt{5}) + 2 (0 - 2) - 32 = 0 [∵ β = - 2]$

$\Rightarrow α (2 \sqrt{5}) = 36 \Rightarrow α = \frac{36}{2 \sqrt{5}}$

$∴$ $α^{2} - β^{2} = {(\frac{36}{2 \sqrt{5}})}^{2} - {(- 2)}^{2} = \frac{1296}{20} - 4 = \frac{1216}{20} = \frac{304}{5}$

Solution

1 $\frac{314}{5}$

2 61

3 $\frac{304}{5}$

4 60

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