If the line l x-m y-1 is a normal to the hyperbola x2-a2-y2-b2-1 then a2-12-b2-m2A- a2-b22B- a2-b2C- a2-b2D- a2-b22

The correct option is D lx + my = 1 is a normal to x^2/a^2+y^2/b^2=1 is a^2/l^2+b^2/m^2a^2/l^2 b^2/m^2=(a^2+b^2)^2/( 1)^2⇒a^2/l^2 b^2/m^2=(a^2+b^2)^2 ...

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Byju's Answer

Standard XII

Physics

Work Done as Dot Product

If the line l...

Question

If the line lx + my = 1 is a normal to the hyperbola $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ =1 then $\frac{a^{2}}{l^{2}} + \frac{b^{2}}{m^{2}}$

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Solution

The correct option is C

lx + my = 1 is a normal to $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ =1 is $\frac{a^{2}}{l^{2}} + \frac{b^{2}}{m^{2}} \frac{a^{2}}{l^{2}} - \frac{b^{2}}{m^{2}} = \frac{(a^{2} + b^{2})^{2}}{(- 1)^{2}} \Rightarrow \frac{a^{2}}{l^{2}} - \frac{b^{2}}{m^{2}} = (a^{2} + b^{2})^{2}$

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Similar questions

Q. If the straight line

l x + m y + n = 0

is normal to the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

then prove that:

\frac{a^{2}}{l^{2}} + \frac{b^{2}}{m^{2}} = \frac{(a^{2} - b^{2})^{2}}{n^{2}}

Q. The condition that the straight line

l x + m y = n

may be a normal to the hyperbola

b^{2} x^{2} - a^{2} y^{2} = a^{2} b^{2}

is given by

Q. If the line

l x + m y + n = 0

is a norma to the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

, then show that

\frac{a^{2}}{l^{2}} - \frac{b^{2}}{m^{2}} = {(\frac{a^{2} + b^{2}}{n})}^{2}

x = $a s e c θ$ + $b t a n θ$

y = $a t a n θ$ + $b s e c θ$

Q. If

P

and

Q

be two points on the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

, whose centre is

C

such that

C P

is perpnediuclar to

C Q, a < b

, then the value of

\frac{1}{C P^{2}} + \frac{1}{C Q^{2}}

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If the line lx + my = 1 is a normal to the hyperbola x2a2+y2b2=1 then a2l2+b2m2

The correct option is C lx + my = 1 is a normal to x2a2+y2b2=1 is a2l2+b2m2a2l2−b2m2=(a2+b2)2(−1)2⇒a2l2−b2m2=(a2+b2)2

If the line lx + my = 1 is a normal to the hyperbola $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ =1 then $\frac{a^{2}}{l^{2}} + \frac{b^{2}}{m^{2}}$