Gujarati
10-2. Parabola, Ellipse, Hyperbola
easy

The point of contact of the tangent $y = x + 2$ to the hyperbola $5{x^2} - 9{y^2} = 45$ is

A

$(9/2, 5/2)$

B

$(5/2, 9/2)$

C

$(-9/2, -5/2)$

D

None of these

Solution

(c) Hyperbola is $\frac{{{x^2}}}{9} – \frac{{{y^2}}}{5} = 1$.

Hence point of contact is $\left[ {\frac{{ – 9(1)}}{{\sqrt {9 – 5} }},\,\frac{{ – 5}}{{\sqrt {9 – 5} }}} \right] \equiv \left[ {\frac{{ – 9}}{2},\,\frac{{ – 5}}{2}} \right]$.

Trick : Since the point $\left( { – \frac{9}{2},\, – \frac{5}{2}} \right)$ satisfies both the equations.

Standard 11
Mathematics

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