First slide

Ellipse

Question

A tangent at any point to the ellipse
$4 x^{2} + 9 y^{2} = 36$ is cut by the tangent at the extremities of the
major axis at $T^{'}$ $T$ and '. The circle on $T T^{'}$ as diameter passes
through the point.

Moderate

A

$(0, \sqrt{5})$
B

$(\sqrt{5}, 0)$
C

$(2, 1)$
D

$(0, - \sqrt{5})$

Solution

Any point on the ellipse is $P (3 \cos θ, 2 \sin θ)$
Equation of the tangent at $P$ is $\frac{x}{3} \cos θ + \frac{y}{2} \sin θ = 1$ which
meets the tangents $x = 3$ and $x = - 3$ at the extremities of
the major axis at
$T (3, \frac{2 (1 - \cos θ)}{\sin θ})$ and $T^{'} (- 3, \frac{2 (1 + \cos θ)}{\sin θ})$
Equation of the circle on $T T^{'}$ as diameter is
$(x - 3) (x + 3) + (y - \frac{2 (1 - \cos θ)}{\sin θ}) (y - \frac{2 (1 + \cos θ)}{\sin θ}) = 0$
$\Rightarrow x^{2} + y^{2} - \frac{4}{\sin θ} y - 5 = 0$ which passes through $(\sqrt{5}, 0)$

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