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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Hyperbola

JEE Main · Mathematics · Hyperbola

If x=9 is a chord of contact of the hyperbola x^2-y^2=9 , then the...

If x=9x=9 is a chord of contact of the hyperbola x2−y2=9x^2-y^2=9, then the equation of the tangent at one of the points of contact is

Options

  1. Ax+3y+2=0x+\sqrt{3} y+2=0
  2. B3x+22y−3=03 x+2 \sqrt{2} y-3=0
  3. C3x−2y+6=03 x-\sqrt{2} y+6=0
  4. Dx−3y+2=0x-\sqrt{3} y+2=0

Correct answer

B. 3x+22y−3=03 x+2 \sqrt{2} y-3=0

Step-by-step solution

Given that, x=9x=9 is a chord of contact of hyperbola. x2−y2=9 put x=9,81−y2=9⇒y2=72⇒y=62 or −62\begin{aligned}& x²-y²=9 \\ & \text { put }x=9, \quad 81-y²=9 \\ & \Rightarrow \quad y²=72 \\ & \Rightarrow \quad y=6 \sqrt{2} \text { or }-6 \sqrt{2}\\ & \end{aligned}\thereforePointsarePoints are(9,6 \sqrt{2})andand(9,-6 \sqrt{2})Now,differentiatingEq.(i)w.r.t.Now, differentiating Eq. (i) w.r.t.x,weget, we get2x−2ydydx=0⇒dydx=xy at (9,62)(dydx)(9,62)=962=322 and at (9−62)(dydx)(9,−62)=−322\begin{aligned}& 2 x-2 y \frac{d y}{d x}=0 \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{x}{y}\\ & \text { at }(9,6 \sqrt{2}) \quad\left(\frac{d y}{d x}\right)_{(9,6 \sqrt{2})}=\frac{9}{6 \sqrt{2}}=\frac{3}{2 \sqrt{2}}\\ & \text { and at }(9-6 \sqrt{2})\left(\frac{d y}{d x}\right)_{(9,-6 \sqrt{2})}=-\frac{3}{2 \sqrt{2}}\\ & \end{aligned}∴\therefore Equation of tangent at (9,62)(9,6 \sqrt{2}) is (y−62)=322(x−9)⇒22y−24=3x−27⇒3x−22y−3=0\begin{aligned} (y-6 \sqrt{2})& =\frac{3}{2 \sqrt{2}}(x-9) \\ \Rightarrow \quad 2 \sqrt{2}y-24 & =3 x-27 \\ \Rightarrow \quad 3 x-2 \sqrt{2}y-3 & =0 \end{aligned}and equation of tangent at (9,−62)(9,-6 \sqrt{2}) is (y+62)=−322(x−9)⇒22y+24=−3x+27⇒3x+22y−3=0\begin{aligned} (y+6 \sqrt{2})& =\frac{-3}{2 \sqrt{2}}(x-9) \\ \Rightarrow \quad 2 \sqrt{2}y+24 & =-3 x+27 \\ \Rightarrow \quad 3 x+2 \sqrt{2}y-3 & =0 \end{aligned}
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