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Parabola question

2018 · 15 Apr · Shift 2 · Q34

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Parabola
2018 · 15 Apr · Shift 2 · Q34

Parabola question

2018 · 15 Apr · Shift 2 · Q34

JEE MainMathematicsParabolaMCQ+4 / −1

Tangents drawn from the point (

-

8, 0) to the parabola y2 = 8x touch the parabola at

P

and

Q.

If F is the focus of the parabola, then the area of the triangle PFQ (in sq. units) is equal to :

A
24
B
32
C
48
D
64

View written solutionFree

Correct answer: C

Identify the parabola and its focus

Given parabola: $y^2=8x$

Compare with the standard form: $y^2=4ax$ So, $4a=8 \implies a=2$

Hence the focus is: $F=(a,0)=(2,0)$

Parametric point on the parabola

A general point on the parabola $y^2=4ax$ is: $(at^2,2at)$

Here $a=2$ , so a general point is: $P=(2t^2,4t)$

The tangent at parameter $t$ to $y^2=4ax$ is: $ty=x+at^2$

For $a=2$ , tangent becomes: $ty=x+2t^2$

Tangents from the point $(-8,0)$

Since the tangent passes through $(-8,0)$ , substitute this point into $ty=x+2t^2$

We get: $t(0)=-8+2t^2$ $0=-8+2t^2$ $2t^2=8$ $t^2=4$ $t=\pm 2$

So the points of contact are:

For $t=2$ : $P=(2\cdot 2^2,4\cdot 2)=(8,8)$
For $t=-2$ : $Q=(2\cdot (-2)^2,4\cdot (-2))=(8,-8)$

Coordinates of triangle vertices

Thus, $P=(8,8), \quad Q=(8,-8), \quad F=(2,0)$

Notice that $PQ$ is a vertical segment.

Length of base $PQ$ : $PQ=8-(-8)=16$

Distance of focus $F(2,0)$ from the line $x=8$ is: $8-2=6$

Area of triangle $PFQ$

Using $\text{Area} = \frac{1}{2}\times \text{base} \times \text{height}$

$\text{Area} = \frac{1}{2}\times 16 \times 6 = 48$

Final answer

$\boxed{48}$

So the correct option is C.

Comparison with stored answer

Stored correct answer: C

My derived answer: C (48)

They match.

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