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Question

Mathematics Question on Application of derivatives

If the tangents drawn to the hyperbola 4y2=x2+14y^2 = x^2 + 1 intersect the co-ordinate axes at the distinct points AA and BB , then the locus of the mid point of ABAB is :

A

x24y2+16x2y2=0x^2 - 4y^2 + 16x^2y^2 = 0

B

x24y216x2y2=0x^2 - 4y^2 - 16x^2y^2 = 0

C

4x2y2+16x2y2=04x^2 - y^2 + 16x^2y^2 = 0

D

4x2y216x2y2=04x^2 - y^2 - 16x^2y^2 = 0

Answer

x24y216x2y2=0x^2 - 4y^2 - 16x^2y^2 = 0

Explanation

Solution

4y2=x2+14 y^{2}=x^{2}+1
Point 4yy1=xx1+1 4 y y_{1}=x x_{1}+1 with 4y12=x12+14 y_{1}^{2}=x_{1}^{2}+1
x axis 1x,0]\frac{-1}{x}, 0]
y axis [0,14y1]\left[0, \frac{1}{4 y_{1}}\right]
Mid point h=12x,k=18y1h=\frac{-1}{2 x}, k=\frac{1}{8 y_{1}}
x1=12hy1=18kx_{1}=\frac{-1}{2 h} y_{1}=\frac{1}{8 k}
4(18k)2=(12h)2+14\left(\frac{1}{8 k}\right)^{2}=\left(\frac{-1}{2 h}\right)^{2}+1
44k2=14b5+1\frac{4}{4 k^{2}}=\frac{1}{4 b^{5}}+1
116y2=14b2+1\frac{1}{16 y^{2}}=\frac{1}{4 b^{2}}+1
116y2=1+4x24x2\frac{1}{16 y^{2}}=\frac{1+4 x^{2}}{4 x^{2}}