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  • If a hyperbola passes through the point P(10,16) and it has vertices at  then the equation of the normal to it at P is : Option: 1</

If a hyperbola passes through the point P(10,16) and it has vertices at (\pm 6,0), then the equation of the normal to it at P is :
Option: 1 3x+4y=94
Option: 2 x+2y=42
Option: 3 2x+5y=100
Option: 4 x+3y=58
 

Answers (1)

best_answer

 

 

What is Hyperbola? -

Hyperbola:

 

The standard form of the equation of a hyperbola with center (0, 0) and major axis on the x-axis is

 

\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1 \quad \text { where, } \mathrm{b}^{2}=\mathrm{a}^{2}\left(\mathrm{e}^{2}-1\right)

 

Important Terms related to Hyperbola:

  • Centre: All chord passing through point O is bisected at point O. Here O is the origin, i.e. (0, 0). 

  • Foci: Point S and S’ is foci of the hyperbola where, S is (ae, 0) and S’ is (-ae, 0).

  • Directrices: The straight line ZM and Z’M’ are two directrices of the hyperbola and their equations are x=ae and x=-ae.

  • Axis: In figure AA’ is called the transverse axis and the line perpendicular it through centre of hyperbola is called  conjugate axis. 2a is length of transverse axis and 2b is length of conjugate axis.

  • Double Ordinate: If a line perpendicular to axis of hyperbola meets the curve at Q and Q’, then QQ’ is called double ordinate. 

  • Latusrectum: Double ordinate passing through focus is called latus rectum. Here LL’ is latusrectum.

 

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Equation of Normal of Hyperbola in Point Form -

Equation of Normal of Hyperbola in Point Form

Point form:

\\ {\text {The equation of normal at }\left(x_{1}, y_{1}\right) \text { to the hyperbola, } \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \text { is }} \\ {\frac{a^{2} x}{x_{1}}+\frac{b^{2} y}{y_{1}}=a^{2}+b^{2} \text { . }}

 

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Vertex is at (±6, 0)  

\thereforea = 6

Let the hyperbola is \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1

Putting point P(10, 16) on the hyperbola

\\ {\frac{100}{36}-\frac{256}{b^{2}}=1} {\Rightarrow b^{2}=144} \\ {\therefore \text { hyperbola is } \frac{x^{2}}{36}-\frac{y^{2}}{144}=1}

\begin{array}{l}{\therefore \text { equation of normal is } \frac{a^{2} x}{x_{1}}+\frac{b^{2} y}{y_{1}}=a^{2}+b^{2}} \\ {\therefore \text { putting we get } 2 x+5 y=100}\end{array}

Correct Option (3)

Posted by

vishal kumar

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