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Byju's Answer

Standard XII

Mathematics

Slope Form of Tangent: Ellipse

Suppose y =...

Question

Suppose $y = m x + 2 \sqrt{5}$ is a tangent to the hyperbola $4 x^{2} - y^{2} = 16 .$ If $m_{1}, m_{2}$ are the two values of $m$ for which this happens, then value of $m_{1} m_{2} + (\frac{m_{1}}{m_{2}})$ is

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Solution

$\begin{matrix} 4 x^{2} - y^{2} = 16 y = m x + 2 \sqrt{5} 4 x^{2} - {(m x + 2 \sqrt{5})}^{2} = 16 4 x^{2} - [m^{2} x^{2} + 20 + 4 \sqrt{5} m x] = 16 (4 - m^{2}) x^{2} - 4 \sqrt{5} m x - 36 = 0 D = 0 s o, m^{2} + 4 (4 - m^{2}) \times 36 = 0 20 m^{2} + (4 - m^{2}) \times 36 = 0 5 m^{2} + 9 (4 - m^{2}) = 0 36 = 4 m^{2} m^{2} = 9 m = \pm 3 n o w, m_{1} m_{2} + (\frac{m_{1}}{m_{2}}) = 3 \times (- 3) + \frac{3}{- 3} = - 10. \end{matrix}$

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Suppose y=mx+2√5 is a tangent to the hyperbola 4x2−y2=16. If m1,m2 are the two values of m for which this happens, then value of m1m2+(m1m2) is

4x2−y2=16y=mx+2√54x2−(mx+2√5)2=164x2−[m2x2+20+4√5mx]=16(4−m2)x2−4√5mx−36=0D=0so,m2+4(4−m2)×36=020m2+(4−m2)×36=05m2+9(4−m2)=036=4m2m2=9m=±3now,m1m2+(m1m2)=3×(−3)+3−3=−10.

Suppose $y = m x + 2 \sqrt{5}$ is a tangent to the hyperbola $4 x^{2} - y^{2} = 16 .$ If $m_{1}, m_{2}$ are the two values of $m$ for which this happens, then value of $m_{1} m_{2} + (\frac{m_{1}}{m_{2}})$ is