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Question: If m<sub>1</sub> and m<sub>2</sub> are the gradients of tangents to hyperbola \(\frac{x^{2}}{25} - \...

If m1 and m2 are the gradients of tangents to hyperbola x225y216=1\frac{x^{2}}{25} - \frac{y^{2}}{16} = 1 which passes through (6, 2), then-

A

m1 + m­2 = 42/11

B

m1m2 = 20/11

C

m1+ m2 = 48/11

D

m1m2 = 11/20

Answer

m1m2 = 20/11

Explanation

Solution

x225\frac{x^{2}}{25}y216\frac{y^{2}}{16} = 1

equation of tangent in slope

y = mx ± 25m216\sqrt{25m^{2} - 16}

It passes through (6, 2)

\ 2 = 6m ± 25m2 – 16

(2 – 6m)2 = 25m2 – 16

4 + 36m2 – 24m = 25m2 – 16

11m2 – 24m + 20 = 0

m1 + m2 = 24/11, m1m2 = 20/11