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Byju's Answer

Standard XII

Mathematics

General Equation of Hyperbola

For a parabol...

Question

For a parabola, if $L_{1} : x = y + 1$ is the axis of symmetry, $L_{2} : x + y = 5$ is tangent at vertex and $L_{3} : y = 4$ is a tangent at a point $P,$ then which of the following is (are) CORRECT ?

Focus of the parabola is

(1, 0)

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The equation of circumcircle of the triangle formed by the tangent and normal at point

P

and axis of parabola is

x^{2} + y^{2} - 2 x = 31

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The length of latus rectum of

y^{2} = 8 \sqrt{2} x

and the given parabola is equal

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The area of the quadrilateral formed by the tangents and normals at the extremities of the latus rectum of the given parabola is

64

sq. units

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Solution

The correct option is D The area of the quadrilateral formed by the tangents and normals at the extremities of the latus rectum of the given parabola is $64$ sq. units
$L_{1} : x = y + 1$ is axis of symmetry.
$L_{2} : x + y = 5$ is tangent at vertex.
$L_{3} : y = 4$ is tangent at $P$
Vertex is given by point of intersection of $L_{1}$ and $L_{2}$
$2 y + 1 = 5 \Rightarrow y = 2 \Rightarrow x = 3 Vertex = V = (3, 2)$

As focus lies on $L_{1},$ let it be $(h, h - 1)$
Foot of perpendicular from focus on $L_{3}$ lies on the tangent at vertex $L_{2}$ , so the point of intersection of $L_{2}$ and $L_{3}$ is $(1, 4)$
Slope of line joining $(h, h - 1)$ and $(1, 4)$ tends to infinity,
$m = \frac{h - 5}{h - 1} ∴ h = 1 Focus = S = (1, 0)$
We know that $V$ is midpoint of $S$ and $A$ , so
$A = (5, 4)$
Directrix is parallel to $L_{2}$ and passing through $A$ , so its equation is
$x + y = 9$
Let point $P = (k, 4)$
Using the definition of parabola,
$S P = P M \Rightarrow \sqrt{(k - 1)^{2} + 16} = ∣ ∣ ∣ \frac{k + 4 - 9}{\sqrt{2}} ∣ ∣ ∣ \Rightarrow 2 (k^{2} - 2 k + 17) = (k^{2} - 10 k + 25) \Rightarrow k^{2} + 6 k + 9 = 0 \Rightarrow k = - 3 ∴ P = (- 3, 4)$
Normal at $P$
$x = - 3$

$B = (- 3, - 4)$
Equation of circumcircle of triangle $P A B$ , where $A B$ is diameter, is
$(x - 5) (x + 3) + (y - 4) (y + 4) = 0 \Rightarrow x^{2} + y^{2} - 2 x = 31$

Length of latus rectum $= 4 S V = 4 \sqrt{2^{2} + 2^{2}} = 8 \sqrt{2}$
$y^{2} = 8 \sqrt{2} x$
Length of latus rectum $= 8 \sqrt{2}$
So, they have same latus rectum.

The quadrilateral formed by tangents and normals at the endpoints of the latus rectum is a square whose diagonal is latus rectum
Area of square $= \frac{{(Diagonal)}^{2}}{2}$
$= \frac{(8 \sqrt{2})^{2}}{2} = 64 sq. units$

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Similar questions

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L_{2} : x + y = 5

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For a parabola, if L1:x=y+1 is the axis of symmetry, L2:x+y=5 is tangent at vertex and L3:y=4 is a tangent at a point P, then which of the following is (are) CORRECT ?

For a parabola, if $L_{1} : x = y + 1$ is the axis of symmetry, $L_{2} : x + y = 5$ is tangent at vertex and $L_{3} : y = 4$ is a tangent at a point $P,$ then which of the following is (are) CORRECT ?