Hyperbola Ques 19
- The equation of a common tangent to the curves, $y^{2}=16 x$ and $x y=-4$, is
(2019 Main, 12 April II)
(a) $x-y+4=0$
(b) $x+y+4=0$
(c) $x-2 y+16=0$
(d) $2 x-y+2=0$
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Answer:
Correct Answer: 19.(a)
Solution:
Formula:
Key Idea An equation of tangent having slope
’ $m$ ’ to parabola $y^{2}=4 a x$ is $y=m x+\frac{a}{m}$.
Given equation of curves are
$$ y^{2}=16 x(\text { parabola }) $$
and $\quad x y=-4$ (rectangular hyperbola)
Clearly, equation of tangent having slope ’ $m$ ’ to parabola
(i) is $y=m x+\frac{4}{m}$
Now, eliminating $y$ from Eqs. (ii) and (iii), we get
$$ x \quad m x+\frac{4}{m}=-4 \Rightarrow m x^{2}+\frac{4}{m} x+4=0 $$
which will give the points of intersection of tangent and rectangular hyperbola.
Since, line $y=m x+\frac{4}{m}$ is also a tangent to the rectangular hyperbola.
$\therefore$ Discriminant of quadratic equation $m x^{2}+\frac{4}{m} x+4=0$, should be zero.
[ $\because$ there will be only one point of intersection]
$$ \begin{array}{ll} \Rightarrow & D=\frac{4}{m}^{2}-4(m)(4)=0 \\ \Rightarrow & m^{3}=1 \Rightarrow m=1 \end{array} $$
So, equation of required tangent is $y=x+4$.
BETA
