Magnitude of gradient of tangent at an extremity of latera recta of hyperbola frac{{{x^2}}}{{{

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10-2. Parabola, Ellipse, Hyperbola

normal

The magnitude of the gradient of the tangent at an extremity of latera recta of the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ is equal to (where $e$ is the eccentricity of the hyperbola)

$be$

$e$

$ab$

$ae$

Solution

Let a hyperbola, $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1-\ldots$

Let the Eccentricity be e. End point or extremities of LR are $\left(\right.$ae, $\left.\frac{\pm b^{2}}{a}\right)$ Tangent at $L \left(\right.$ ae $\left., \frac{b^{2}}{a}\right)$ will be

$\Rightarrow \frac{x(a e)}{a^{2}}-\frac{y\left(\frac{b^{2}}{a}\right)}{b^{2}}=1\left(\right.$ as tangent at $\left(x_{1}, y_{1}\right)$ is $\left.\frac{x x_{1}}{a^{2}}-\frac{y y_{1}}{b^{2}}=1\right)$

$\Rightarrow \frac{x e}{a}-\frac{y}{a}=1$

$\Rightarrow x e-y=a \Rightarrow y=x e-a-x-2$

General equation of a tangent in slope form $y=m x+c$ –

Comparing Equation $2 \& 3$, as they represent the tangents of Hyperbola,

$m=e$

The gradient i.e. slope of tangent $=m=e$

Standard 11

Mathematics

The eccentricity of the conic ${x^2} – 4{y^2} = 1$, is

easy

$P$ is a point on the hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}}$ $= 1, N $ is the foot of the perpendicular from $P$ on the transverse axis. The tangent to the hyperbola at $P$ meets the transverse axis at $ T$ . If $O$ is the centre of the hyperbola, the $OT. ON$ is equal to :

normal

Eccentricity of the rectangular hyperbola $\int_0^1 {{e^x}\left( {\frac{1}{x} – \frac{1}{{{x^3}}}} \right)} \;dx$ is

easy

If the tangents drawn to the hyperbola $4y^2 = x^2 + 1$ intersect the co-ordinate axes at the distinct points $A$ and $B$, then the locus of the mid point of $AB$ is

hard

(JEE MAIN-2018)

If the line $y = 2x + \lambda $ be a tangent to the hyperbola $36{x^2} – 25{y^2} = 3600$, then $\lambda = $

easy

The magnitude of the gradient of the tangent at an extremity of latera recta of the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ is equal to (where $e$ is the eccentricity of the hyperbola)

Solution

Similar Questions

The eccentricity of the conic ${x^2} – 4{y^2} = 1$, is

$P$ is a point on the hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}}$ $= 1, N $ is the foot of the perpendicular from $P$ on the transverse axis. The tangent to the hyperbola at $P$ meets the transverse axis at $ T$ . If $O$ is the centre of the hyperbola, the $OT. ON$ is equal to :

Eccentricity of the rectangular hyperbola $\int_0^1 {{e^x}\left( {\frac{1}{x} – \frac{1}{{{x^3}}}} \right)} \;dx$ is

If the tangents drawn to the hyperbola $4y^2 = x^2 + 1$ intersect the co-ordinate axes at the distinct points $A$ and $B$, then the locus of the mid point of $AB$ is

If the line $y = 2x + \lambda $ be a tangent to the hyperbola $36{x^2} – 25{y^2} = 3600$, then $\lambda = $

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