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Position of a Point W.R.T Hyperbola

Let t 1, t 2,...

Question

Let $t_{1}, t_{2}, t_{3}$ be three points on the parabola $y^{2} = 4 x$ . If the normal at $t_{1}$ intersects the parabola at $t_{2}$ and the normal at $t_{2}$ intersects the parabola at $t_{3}$ such that $3 t_{1} + 13 t_{2} + 9 t_{3} = 0$ , then

t_{1} = 1

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t_{1} = 2

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t_{2} = 3

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t_{3} = \frac{1}{3}

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Solution

The correct options are
B $t_{1} = 2$
D $t_{3} = \frac{1}{3}$
$t_{2} = - t_{1} - \frac{2}{t_{1}}$
$t_{3} = - t_{2} - \frac{2}{t_{2}}$
and $3 t_{1} + 13 t_{2} + 9 t_{3} = 0$
$\Rightarrow t_{1} = 2, t_{2} = - 3, t_{3} = \frac{11}{3}$

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Similar questions

Q. If

t_{1}

t_{2}

t_{3}

are three points on

y^{2} = 4 x

such that normal at

t_{1}

intersects the parabola at

t_{2}

and normal at

t_{2}

intersects the parabola at

t_{3}

and

3 t_{1} + 13 t_{2} + 9 t_{3}

then

t_{1}

The area of the triangle formed by three points $P {a t_{1} t_{2}, a (t_{1} + t_{2})}, Q {a t_{2} t_{3}, a (t_{2} + t_{3})}, R {a t_{3} t_{1}, a (t_{3} + t_{1})}$ is _____.

Q. If a circle intersects the parabola

y^{2} = 4 a x

at point

A (a t_{1}^{2}, 2 a t_{1}), B (a t_{2}^{2}, 2 a t_{2}), C (a t_{3}^{2}, 2 a t_{3}), D (a t_{4}^{2}, 2 a t_{4})

, then

t_{1} + t_{2} + t_{3} + t_{4} + t_{4}

Let $θ \in (0, \frac{π}{4}) a n d t_{1} = (t a n θ)^{t a n θ}, t_{2} (t a n θ)^{c o t} t_{3} = (c o t θ)^{t a n θ} a n d t_{4} (c o t θ)^{t a n θ}$ , then

I, I I, I I I

are three isotherms respectively at

T_{1}, T_{2}

and

T_{3}

. The temperatures will be in the order:

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Let t1,t2,t3 be three points on the parabola y2=4x. If the normal at t1 intersects the parabola at t2 and the normal at t2 intersects the parabola at t3 such that 3t1+13t2+9t3=0, then

The correct options are B t1=2 D t3=13t2=−t1−2t1 t3=−t2−2t2 and 3t1+13t2+9t3=0 ⇒t1=2,t2=−3,t3=113

Let $t_{1}, t_{2}, t_{3}$ be three points on the parabola $y^{2} = 4 x$ . If the normal at $t_{1}$ intersects the parabola at $t_{2}$ and the normal at $t_{2}$ intersects the parabola at $t_{3}$ such that $3 t_{1} + 13 t_{2} + 9 t_{3} = 0$ , then

The correct options are
B $t_{1} = 2$
D $t_{3} = \frac{1}{3}$
$t_{2} = - t_{1} - \frac{2}{t_{1}}$
$t_{3} = - t_{2} - \frac{2}{t_{2}}$
and $3 t_{1} + 13 t_{2} + 9 t_{3} = 0$
$\Rightarrow t_{1} = 2, t_{2} = - 3, t_{3} = \frac{11}{3}$