Question

Consider the parabola $y^2=8x$. Let ${△}_1$ be the area of the triangle formed by the end points of its latus rectum and the point $P(\frac{1}{2},2)$ on the parabola, and ${△}_2$ be the area of the triangle formed by drawing tangents at $P$ and at the end points of the latus rectum. Then $\frac{{△}_1}{{△}_2}=$

• A. 02
• B. 2.00
• C. 2.0
• D. 2

Answer 1

Answer: (A), (B), (C), (D)

Method 1:
Given parabola is $y^2=8x$
Comparing with $y^2=4ax,$ we get $a=2$
So directrix is $x=-2$
Tangents at end points of latus rectum meet on directrix on $x$ axis.
So that point will be $(-2,0)$

Area of triangle formed by three points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by:
$\frac{1}{2}|[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]|$

${△}_1=\text{Area of}△ABC$
On substituting the points $A(2,4),B(2,-4)\&C(-2,0)$ in the above equation, we will get
$\Rightarrow \frac{1}{2}|2(-4-0)+2(0-4)-2(4+4)=\frac{12}{2}=6$
Now ${△}_2=\text{Area of}△CQR$
Tangent at a point $(x_1,y_1)$ is $y{y}_1=2a(x+x_1)$
Now, tangent at point $A(2,4)$ is $4y=4(x+2)\Rightarrow y=x+\mathrm{2......}\left(i\right)$
Tangent at point $B(2,-4)$ is $-4y=4(x+2)\Rightarrow y=-x-\mathrm{2......}\left(ii\right)$
Also, tangent at point $P(\frac{1}{2},2)$ is $2y=4(x+\frac{1}{2})\Rightarrow y=2x+\mathrm{1......}\left(iii\right)$

Solving $\left(i\right)\&\left(iii\right)$ we will get $Q$
$\Rightarrow 2(x+2)=4(x+\frac{1}{2})\Rightarrow x+2=2x+1\Rightarrow x=1$
$\Rightarrow y=3$

Solving $\left(ii\right)\&\left(iii\right)$ we will get $R$
$\Rightarrow -x-2=2x+1\Rightarrow 3x=-3\Rightarrow x=-1$
$\Rightarrow y=-1$

On substituting the points $C(-2,0),Q(1,3)\&R(-1,-1)$ in the equation of triangle, we will get
$\therefore$ Area of $△CQR$ is
$\Rightarrow \frac{1}{2}|-2(3+1)+1(-1-0)-1(0-3)=\frac{6}{2}=3$

Hence the ratio of area is $\frac{6}{3}=2$

Method 2:
We know that the area of triangle formed by 3 points on parabola = 2(area of triangle formed by tangents)
Here we can see from the figure that ${△}_1$ is the area of triangle formed by 3 points on parabola and ${△}_2$ is the area of triangle formed by the respective tangents.
$\Rightarrow \frac{{△}_1}{{△}_2}=2$.