The tangent drawn at any point of the curve -x-y-a meets the axes in A and B- Prove that OA-OB is a constant-

Lets differentiate √(x)+√(y)=√(a)....(1) wrt to x . 12√(x)+12√(y)dydx=0 dydx= √(y)√(x) .......... (2) Lets assume that the tangent has t ...

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

Byju's Answer

Standard XII

Mathematics

Point of Inflection

The tangent d...

Question

The tangent drawn at any point of the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ meets the axes in A and B. Prove that $O A + O B$ is a constant.

Open in App

Solution

Lets differentiate $\sqrt{x} + \sqrt{y} = \sqrt{a} . . . . (1)$ wrt to $x$ .

$\frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = 0$
$⟹ \frac{d y}{d x} = - \frac{\sqrt{y}}{\sqrt{x}}$ .......... $(2)$

Lets assume that the tangent has the equation in this form,
$\frac{x}{O A} + \frac{y}{O B} = 1$
$⟹$ slope $= - \frac{O B}{O A}$

From $(2)$ ,
$- \frac{\sqrt{y}}{\sqrt{x}} = - \frac{O B}{O A}$
$⟹ \sqrt{y} = c \times O B$ & $\sqrt{x} = c \times O A$ where $c$ is a constant.

$∴ O A + O B = c (\sqrt{x} + \sqrt{y})$
$⟹ O A + O B = c \sqrt{a}$ ........ $(3)$ from $(1)$
$⟹ O A + O B =$ constant.

Hence Proved.

Suggest Corrections

Similar questions

Q. If the tangent at any point of the curve

\sqrt{x} + \sqrt{y} = \sqrt{a}

meets the co-ordinate axes of A and B then OA+OB=

Q. lf normal is drawn at a point P (x, y) of a curve meets the X-axis and the Y- axis in point A and B respectively such that

\frac{1}{O A} + \frac{1}{O B} = 1

, (where '0' is the origin). The equation of the curve is

Q. Assertion :A normal is drawn at a point

P (x, y)

of a curve. It meets the x-axis and the y-axis in point

A

and

B

, respectively, such that

\frac{1}{O A} + \frac{1}{O B} = 1

, where

O

is the origin. The equation of such a curve passing through

(5, 4)

{(x - 1)}^{2} + {(y - 1)}^{2} = 25

Reason:

O A = x + y \frac{d y}{d x}

and

O B = \frac{x + y \frac{d y}{d x}}{\frac{d y}{d x}}

Q. Assertion(A): If the tangent at any point

P

on the curve

x y = a^{2}

meets the axes at

A

and

B

then

A P : P B = 1 : 1

Reason(R): The tangent at

P (x, y)

on the curve

X^{m} . Y^{n} = a^{m + n}

meets the axes at

A

and

B

. Then the ratio of

P

divides

¯ ¯¯¯¯¯¯ ¯ A B

n : m

Q. The line

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

cust the axes in

A

and

B

. Another variable line cuts at the axes in

A^{'}

and

B^{'}

such that

O A + O B = O A^{'} + O B^{'}

then prove that the locus of the point of intersection of the lines

A B^{'}

and

A^{'} B

is the line

x + y = a + b

Join BYJU'S Learning Program

The tangent drawn at any point of the curve √x+√y=√a meets the axes in A and B. Prove that OA+OB is a constant.

The tangent drawn at any point of the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ meets the axes in A and B. Prove that $O A + O B$ is a constant.