Question

For what value of m, is the equation y - x + 3 = 0 a tangent to the curve y² + x² = m².​

Answer 1

Step-by-step explanation:

Given:

$y - x + 3 = 0 \\ {y}^{2} + {x}^{2} = {m}^{2}$

To find: For what value of m, is the equation y - x + 3 = 0 a tangent to the curve y² + x² = m².

Solution:

Step 1: Put value of y from line in given curve

${(x - 3)}^{2} + {x}^{2} = {m}^{2} \\ \\ {x}^{2} - 6x + 9 + {x}^{2} - {m}^{2} = 0 \\ \\ 2 {x}^{2} - 6x + 9 - {m}^{2} = 0$

Step 2 : Find the value of discrimination of quadratic equation and put it to zero, where the tangent touches the curve discriminate should equal to zero.

D=b²-4ac

here,

a=2

b=-6

c=9-m²

$D = ( { - 6)}^{2} - 4(2)(9 - {m}^{2} ) \\ \\Put\:D=0\\\\ 36 - 72 + 8 {m}^{2} = 0 \\ \\ 8 {m}^{2} = 36 \\ \\ {m}^{2} = \frac{36}{8} \\ \\ {m}^{2} = \frac{9}{2} \\ \\ m = ± \frac{3}{ \sqrt{2} }$

Final answer:

Value of m is ±3/√2.

Hope it helps you.

To learn more on brainly:

The line 3y=4x -15 intersects the curve 8x2 = 45 + 27y2 at the points A & B. Find the coordinates of A and B.

https://brainly.in/question/45357530

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:y - x + 3 = 0 \: is \: tangent \: to \: curve \: {x}^{2} + {y}^{2} = {m}^{2}$

Let assume that the point of contact of tangent with given curve is (h, k).

So, (h, k) lies on the

$\rm :\longmapsto\:y - x + 3 = 0 \: and \: {x}^{2} + {y}^{2} = {m}^{2}$

Thus,

$\rm :\longmapsto\:k - h + 3 = 0 - - - (1) \:$

and

$\rm :\longmapsto\: {h}^{2} + {k}^{2} = {m}^{2} - - - - (2)$

Now, Consider

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = {m}^{2}$

On differentiating both sides w. r. t. x we get

$\rm :\longmapsto\:\dfrac{d}{dx}{x}^{2} + \dfrac{d}{dx}{y}^{2} = \dfrac{d}{dx} {m}^{2}$

$\rm :\longmapsto\:2x + 2y\dfrac{dy}{dx} = 0$

$\bf\implies \:\dfrac{dy}{dx} = - \dfrac{y}{x}$

So, slope of tangent at (h, k) is

$\bf\implies \:\dfrac{dy}{dx}_{(h,k)} = - \: \dfrac{k}{h}$

As equation of tangent is

$\rm :\longmapsto\:y - x + 3 = 0$

$\rm :\longmapsto\:y = x - 3$

So, slope of tangent = 1

Hence,

$\rm :\longmapsto\: - \: \dfrac{k}{h} = 1$

$\bf\implies \:h \: = - \: k - - - - (3)$

As, from equation (1),

$\bf\implies \:k - h \: = - 3$

$\bf\implies \:k + k \: = - 3$

$\bf\implies \:2k \: = - 3$

$\bf\implies \:k \: = - \dfrac{3}{2}$

So,

$\bf\implies \:h \: = \: \dfrac{3}{2}$

On substituting the values of h and k, in equation (2), we get

$\rm :\longmapsto\: {\bigg[\dfrac{ - 3}{2} \bigg]}^{2} + {\bigg[\dfrac{3}{2} \bigg]}^{2} = {m}^{2}$

$\rm :\longmapsto\: {\bigg[\dfrac{3}{2} \bigg]}^{2} + {\bigg[\dfrac{3}{2} \bigg]}^{2} = {m}^{2}$

$\rm :\longmapsto\: 2{\bigg[\dfrac{3}{2} \bigg]}^{2} = {m}^{2}$

$\bf\implies \:m \: = \: \pm \: \dfrac{3}{2} \sqrt{2}$