If m1 and m2 are the slopes of the tangents to the hyperbola x225-y216-1 which passes through the point 6-2- then

The correct options are A m_1+m_2=2411 B m_1m_2=2011 Suppose the tangent y=mx+√(25m^2 16) passes through the point #160; (6,2). ⇒ 2= ...

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Byju's Answer

Standard XII

Mathematics

Point Form of Tangent: Ellipse

If m1 and ...

Question

If $m_{1}$ and $m_{2}$ are the slopes of the tangents to the hyperbola $\frac{x^{2}}{25} - \frac{y^{2}}{16} = 1$ which passes through the point $(6, 2)$ , then

m_{1} m_{2} = \frac{20}{11}

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m_{1} + m_{2} = \frac{24}{11}

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m_{1} + m_{2} = \frac{48}{11}

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m_{1} m_{2} = \frac{11}{20}

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Solution

The correct options are
A $m_{1} + m_{2} = \frac{24}{11}$
B $m_{1} m_{2} = \frac{20}{11}$
Suppose the tangent $y = m x + \sqrt{25 m^{2} - 16}$ passes through the point $(6, 2) .$

$\Rightarrow 2 = 6 m + \sqrt{25 m^{2} - 16}$

$\Rightarrow 2 - 6 m = \sqrt{25 m^{2} - 16}$

Squaring both sides, we get

$\Rightarrow {(2 - 6 m)}^{2} = 25 m^{2} - 16$

$\Rightarrow 4 + 36 m^{2} - 24 m = 25 m^{2} - 16$

$\Rightarrow 4 + 36 m^{2} - 24 m - 25 m^{2} + 16 = 0$

$\Rightarrow 11 m^{2} - 24 m + 20 = 0$

Let $m_{1}$ and $m_{2}$ be the roots of the above quadratic equation,

$\Rightarrow$ Sum of the roots $= m_{1} + m_{2} = - \frac{- 24}{11} = \frac{24}{11}$

And Product of the roots $= m_{1} m_{2} = \frac{20}{11}$

Hence $m_{1} + m_{2} = \frac{24}{11}$ and $m_{1} m_{2} = \frac{20}{11}$

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If m1 and m2 are the slopes of the tangents to the hyperbola x225−y216=1 which passes through the point (6,2), then

If $m_{1}$ and $m_{2}$ are the slopes of the tangents to the hyperbola $\frac{x^{2}}{25} - \frac{y^{2}}{16} = 1$ which passes through the point $(6, 2)$ , then