Show that the curves x 2 a 2- 1- y 2 b 2- 1-1 and x 2 a 2- 2- y 2 b 2- 2-1 intersect at right angles-

We #160;have, #160;x2a2+ #955;1+y2b2+ #955;1=1 #160; #160; #160; #160;...1and #160;x2a2+ #955;2+y2b2+ #955;2=1 #160; #160; #160; # ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Angle of Intersection of Two Curves

Show that the...

Question

Show that the curves $\frac{x^{2}}{a^{2} + λ_{1}} + \frac{y^{2}}{b^{2} + λ_{1}} = 1 and \frac{x^{2}}{a^{2} + λ_{2}} + \frac{y^{2}}{b^{2} + λ_{2}} = 1$ intersect at right angles.

Open in App

Solution

$We have, \frac{x^{2}}{a^{2} + λ_{1}} + \frac{y^{2}}{b^{2} + λ_{1}} = 1 . . . (1) and \frac{x^{2}}{a^{2} + λ_{2}} + \frac{y^{2}}{b^{2} + λ_{2}} = 1 . . . (2) Now we can find the slope of both the curve by differentiating w . r . t x \Rightarrow \frac{2 x}{a^{2} + λ_{1}} + \frac{2 y \frac{d y}{d x}}{b^{2} + λ_{1}} = 0 and \frac{2 x}{a^{2} + λ_{2}} + \frac{2 y \frac{d y}{d x}}{b^{2} + λ_{2}} = 0 \Rightarrow \frac{d y}{d x} = - \frac{x}{y} \times \frac{b^{2} + λ_{1}}{a^{2} + λ_{1}} and \frac{d y}{d x} = - \frac{x}{y} \times \frac{b^{2} + λ_{2}}{a^{2} + λ_{2}} \Rightarrow m_{1} = - \frac{x}{y} \times \frac{b^{2} + λ_{1}}{a^{2} + λ_{1}} and m_{2} = - \frac{x}{y} \times \frac{b^{2} + λ_{2}}{a^{2} + λ_{2}} Subtracting (2) from (1), we get, x^{2} (\frac{1}{a^{2} + λ_{1}} - \frac{1}{a^{2} + λ_{2}}) + y^{2} (\frac{1}{b^{2} + λ_{1}} - \frac{1}{b^{2} + λ_{2}}) = 0 \Rightarrow \frac{x^{2}}{y^{2}} = \frac{λ_{2} - λ_{1}}{(b^{2} + λ_{1}) (b^{2} + λ_{2})} \times \frac{1}{\frac{λ_{1} - λ_{2}}{(a^{2} + λ_{1}) (a^{2} + λ_{2})}} N o w, m_{1} \times \times m_{2} = \frac{x^{2}}{y^{2}} \times \frac{b^{2} + λ_{1}}{a^{2} + λ_{1}} \times \frac{b^{2} + λ_{2}}{a^{2} + λ_{2}} = \frac{λ_{2} - λ_{1}}{(b^{2} + λ_{1}) (b^{2} + λ_{2})} \times \frac{(a^{2} + λ_{1}) (a^{2} + λ_{2})}{λ_{1} - λ_{2}} \times \frac{b^{2} + λ_{1}}{a^{2} + λ_{1}} \times \frac{b^{2} + λ_{2}}{a^{2} + λ_{2}} = - 1 hence, (1) and (2) cuts orthogonally .$

Suggest Corrections

Similar questions

Q. Find the condition for the following set of curves to intersect orthogonally:

(i)

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

(ii)

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 and \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1

Q. An ellipse has a focus at

(a e, 0)

and directrix along

x = \frac{a}{e}

. If

b^{2} = a^{2} (1 - e^{2})

, then the equation of ellipse is

If the eccentricities of the hyperbolas $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 a n d \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$ be e and $e_{1}$ , then $\frac{1}{e^{2}} + \frac{1}{e_{1}^{2}} =$

Q. Let a, b, c be positive real numbers. The following system of equations in x, y and z

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1, \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1, - \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1 has

(a) no solution
(b) unique solution
(c) infinitely many solutions
(d) finitely many solutions

Q. The area enclosed by a curve with equation

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

Join BYJU'S Learning Program

Show that the curves x2a2+λ1+y2b2+λ1=1 and x2a2+λ2+y2b2+λ2=1intersect at right angles.

Show that the curves $\frac{x^{2}}{a^{2} + λ_{1}} + \frac{y^{2}}{b^{2} + λ_{1}} = 1 and \frac{x^{2}}{a^{2} + λ_{2}} + \frac{y^{2}}{b^{2} + λ_{2}} = 1$ intersect at right angles.