Let P be the point on the parabola y2 = 4x wh ich is at the shortest distance from the center S of the circle x2 + y2 andndash; 4x andndash;16y + 64 = 0. Let Q be the point on the circle dividing the line segment SP internally. Then (JEE Adv. 2016)a)SP = b)SQ : QP = c)the x-in tercept of the normal to the parabola at P is 6d)the slope of the tangent t o the circle at Q is Correct answer is option 'A,C,D'. Can you explain this answer?

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Let P be the point on the parabola y² = 4x wh ich is at the shortest distance from the center S of the circle x² + y²– 4x –16y + 64 = 0. Let Q be the point on the circle dividing the line segment SP internally. Then (JEE Adv. 2016)

a)
SP =
b)
SQ : QP =
c)
the x-in tercept of the normal to the parabola at P is 6
d)
the slope of the tangent t o the circle at Q is

Correct answer is option 'A,C,D'. Can you explain this answer?

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Let P be the point on the parabola y2 = 4x wh ich is at the shortest d...

Let point P on parabola y² = 4x be (t², 2t)

∵ PS is shortest distance, therefore PS should be the normal to parabola.

Equation of normal to y² = 4x at P (t², 2t) is y – 2t = – t(x – t²)
It passes through S(2, 8)
∴ 8 – 2t = – t(2 – t²) ⇒ t³ = 8 or t = 2
∴ P(4, 4)
Also slope of tangent to circle at

Equation of normal at t = 2 is 2x + y = 12
Clearly x-intercept = 6
SP =

and SQ = r = 2
∴ Q divides SP in the ratio SP : PQ

Hence a, c , d are the correct options.

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Let P be the point on the parabola y2 = 4x wh ich is at the shortest distance from the center S of the circle x2 + y2 – 4x –16y + 64 = 0. Let Q be the point on the circle dividing the line segment SP internally. Then (JEE Adv. 2016)a)SP = b)SQ : QP = c)the x-in tercept of the normal to the parabola at P is 6d)the slope of the tangent t o the circle at Q is Correct answer is option 'A,C,D'. Can you explain this answer?

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Let P be the point on the parabola y2 = 4x wh ich is at the shortest distance from the center S of the circle x2 + y2 – 4x –16y + 64 = 0. Let Q be the point on the circle dividing the line segment SP internally. Then (JEE Adv. 2016)a)SP = b)SQ : QP = c)the x-in tercept of the normal to the parabola at P is 6d)the slope of the tangent t o the circle at Q is Correct answer is option 'A,C,D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let P be the point on the parabola y2 = 4x wh ich is at the shortest distance from the center S of the circle x2 + y2 – 4x –16y + 64 = 0. Let Q be the point on the circle dividing the line segment SP internally. Then (JEE Adv. 2016)a)SP = b)SQ : QP = c)the x-in tercept of the normal to the parabola at P is 6d)the slope of the tangent t o the circle at Q is Correct answer is option 'A,C,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let P be the point on the parabola y2 = 4x wh ich is at the shortest distance from the center S of the circle x2 + y2 – 4x –16y + 64 = 0. Let Q be the point on the circle dividing the line segment SP internally. Then (JEE Adv. 2016)a)SP = b)SQ : QP = c)the x-in tercept of the normal to the parabola at P is 6d)the slope of the tangent t o the circle at Q is Correct answer is option 'A,C,D'. Can you explain this answer?.

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