Question

The tangents and normals at the ends of the latus rectum of a parabola forms a

• A. parallelogram
• B. rectangle
• C. square
• D. rhombus

Answer 1

The correct option is C square


Let parabola be $y^2=4ax$
End points of latus rectum are
$L(a,2a),L^{\prime }(a,-2a)$

Tangents at $L$ and $L^{\prime }$ are
$T=0\Rightarrow y\left(2a\right)=2a(x+a)\Rightarrow y=x+a$
And
$y(-2a)=2a(x+a)\Rightarrow -y=x+a$
Both the tangents intersect at $Z(-a,0)$

Normals at $L(a,2a)$ and $L^{\prime }(a,-2a)$ are
$y=-x+3a\text{and}y=x-3a$
The normals intersect at $N(3a,0)$

By using ditance formula
$LN=2\sqrt{2}a{L}^{\prime }N=2\sqrt{2}aLZ=2\sqrt{2}a{L}^{\prime }Z=2\sqrt{2}a$
Also, the tangent and normal are perpendicular to each other and $LZ\perp L^{\prime }Z,LN\perp L^{\prime }N$
Hence it's a square.