The line y -3-2 x and y -2-5 x intersect the curve 3 x 2-4 xy -5 y 2-4-0 at the points P and Q respectively- The tangents drawn to the curve at P and QA- Intersect each other at angle of 45-B- are parallel to each otherC- Intersect each other at angel of 15-D- are perpendicular to each other

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Definition of Functions

The line y =-...

Question

The line $y = - \frac{3}{2} x a n d y = - \frac{2}{5} x$ intersect the curve $3 x^{2} + 4 x y + 5 y^{2} - 4 = 0$ at the points P and Q respectively. The tangents drawn to the curve at P and Q

Intersect each other at angle of

45^{\circ}

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are parallel to each other

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are perpendicular to each other

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Intersect each other at angel of

15^{\circ}

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Solution

The correct option is C are perpendicular to each other
Given, $3 x^{2} + 4 x y + 5 y^{2} - 4 = 0$ , On differentiating w.r.t.x, we get
$\frac{d y}{d x} = - (\frac{2 y + 3 x}{2 x + 5 y}) \Rightarrow {\frac{d y}{d x}]}_{(x_{1}, y_{1})} = 0 a n d {\frac{d y}{d x}]}_{(x_{2}, y_{2})} = \infty$
$\Rightarrow$ Tangents are perpendicular to each other.

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