# Get Answers to all your Questions

#### Find the locus of points of intersection of tangents drawn at the ends of all normal chords to the parabola $\mathrm{y^2=8(x-1)}$  Option: 1 $\mathrm{y^2(x+6)+32-0}$  Option: 2 $\mathrm{y^2 x+32=0}$  Option: 3 $\mathrm{y^2(x+3)+32=0}$  Option: 4 $\mathrm{y^2(x+5)+32=0}$

Consider the parabola $\mathrm{y^2=4 a x}$ Let $\mathrm{P Q}$  be any normal chord whose equation is
$\mathrm{y=m x-2 a m-a m^3}$      ......(1)
If the tangents at $\mathrm{P}$ and intersect at $\mathrm{(h, k)}$ then $\mathrm{PQ}$is chord of contact whose equation is
$\mathrm{k y=2 a(x+h)}$      ...(2)
Comparing (1) and (2) we get
$\mathrm{ \frac{k}{1}=\frac{2 a}{m}=\frac{2 a h}{-a\left(2 m+m^3\right)} }$
$\mathrm{\therefore \quad m=\frac{2 a}{k}\: and \: k\left(2 m+m^3\right)=-2 h}$
Putting for $\mathrm{m \: we \: get\: k\left(\frac{4 a}{k}+\frac{8 a^3}{k^3}\right)=-2 h}$
$\mathrm{or \: 2 a+\frac{4 a^3}{k^2}=-h}$
$\mathrm{\therefore k^2(h+2 a)+4 a^3=0}$
Hence the required locus is $\mathrm{y^2(x+2 a)+4 a^3=0}$

Here parabola is $\mathrm{y^2=8(x-1)}$

$\mathrm{\therefore a=2 \: \: and\, \, replace\: \: x \: \: by\: \: x-1}$

Hence the locus is $\mathrm{y^2(x-1+2.2)+4 .(2)^3}$

$\mathrm{or \: y^2(x+3)+32=0}$

## JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE