The number of possible tangents which can be drawn to the curve $4 x^{2} - 9 y^{2} = 36$ , which are perpendicular to the straight line $5 x + 2 y - 10 = 0$ is

Zero

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Solution

The correct option is A Zero
The line perpendicular to $5 x + y - 10 = 0$ is $x - 5 y + k = 0$
slope $= \frac{1}{5}$
Slope of tangent at a point to curve is $\frac{d y}{d x} = \frac{4 x}{9 y}$
according to question
$\frac{1}{5} = \frac{4 x}{9 y} ⟹ 20 x - 9 y = 0$ which has no point in common with curve
So there is no such Tangent

Suggest Corrections

Similar questions

4 x^{2} - 9 y^{2} = 36

5 x + 2 y - 10 = 0

4 x^{2} + 9 y^{2} = 1

2 y + x = 0

\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1

5 x + 2 y = 10

\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1

5 x + 2 y = 10

9 y^{2} - 4 x^{2} = 1

3 y = x + 7

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