VIT - VITEEE 2025
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
Position of a point with respect to Hyperbola is considered one of the most asked concept.
11 Questions around this concept.
Let $a$ and $b$ be the semi-transverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation $9 e^2-18 e+5=0$. If $S(5,0)$ is a focus and $5 x=9$ is the corresponding directrix of this hyperbola, then $\mathrm{a}^2-\mathrm{b}^2$ is equal to :
The foci of a hyperbola coincide with the foci of ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$. If the eccentricity of the hyperbola is 3, then its equation is
If the polar of a point w.r.t. touches the hyperbola
, then the locus of the point is:
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The locus of the poles of the chords of the hyperbola , which subtend a right angle at the centre is:
Bag A contains 3 white, 7 red balls and Bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn is white, is
$
\text { The value of } \alpha \text { such that } \alpha(2 \alpha, 3 \alpha) \text { lies inside the hyperbola } \frac{x^2}{16}-\frac{y^2}{9}=1 \text {, lies in the interval }
$
The line $25 \mathrm{x}+12 \mathrm{y}-45=0$ meets the hyperbola $25 \mathrm{x}^2-9 \mathrm{y}^2=225$ only at point $\left(5,-\frac{5 \mathrm{k}}{3}\right)$, then the value of k is
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
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Position of a point concerning Hyperbola
Let P(x1,y1) be any point in the plane
(a) P lies outside of the hyperbola : $\frac{\mathrm{x}_1{ }^2}{\mathrm{a}^2}-\frac{\mathrm{y}_1{ }^2}{\mathrm{~b}^2}-1<0$
(b) P lies on of the hyperbola $\quad: \frac{\mathrm{x}_1{ }^2}{\mathrm{a}^2}-\frac{\mathrm{y}_1{ }^2}{\mathrm{~b}^2}-1=0$
(c) P lies inside of the hyperbola : $\frac{\mathrm{x}_1{ }^2}{\mathrm{a}^2}-\frac{\mathrm{y}_1{ }^2}{\mathrm{~b}^2}-1>0$
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