Question

The equation of the common tangent to the Curves,
$\bf {y}^{2} = 16x \: \: and \: \: xy = - 4 \: \: is$
​

Answer 1

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

Given equation of Parabola is

$\rm :\longmapsto\: {y}^{2} = 16x$

We know,

If m is the slope of tangent to the Parabola is y² = 4ax, then equation of tangent is

$\rm :\longmapsto\:y = mx + \dfrac{a}{m}$

If we compare the given Equation of Parabola y² = 16x with y² = 4ax, we get a = 4.

So,

Equation of tangent having slope 'm' to the parabola y² = 16x is given by

$\rm :\longmapsto\:y = mx + \dfrac{4}{m}$

$\rm :\longmapsto\:y = \dfrac{ {xm}^{2} + 4}{m} - - (1)$

Now, for this line to be tangent to xy = - 4, we have

$\rm :\longmapsto\: \dfrac{ {xm}^{2} + 4}{m} \times x = - 4$

$\rm :\longmapsto\: {x}^{2} {m}^{2} + 4x = - 4m$

$\rm :\longmapsto\: {m}^{2} {x}^{2} + 4x + 4m = 0$

For this line to be tangent, Discriminant = 0

• i.e. b² - 4ac = 0

Here,

• a = m²

• b = 4

• c = 4m

So, on substituting the values, we get

$\rm :\longmapsto\: {4}^{2} - 4 \times 4m \times {m}^{2} = 0$

$\rm :\longmapsto\:16 - 16 {m}^{3} = 0$

$\rm :\longmapsto\:16 {m}^{3} = 16$

$\rm :\longmapsto\: {m}^{3} = 1$

$\bf\implies \:m = 1$

On substituting the value of 'm', in equation (1), we get

$\rm :\longmapsto\:y = \dfrac{ {x}(1)^{2} + 4}{1}$

$\rm :\longmapsto\:y = \dfrac{ {x} + 4}{1}$

$\rm :\longmapsto\:y = x + 4$

Hence,

The equation of common tangent to

$\rm :\longmapsto\: {y}^{2} = 16x$

and

$\rm :\longmapsto\:xy = - 4$

is

$\bf :\longmapsto\:y = x + 4$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}Solution−</p><p></p><p>Given equation of Parabola is</p><p></p><p>\rm :\longmapsto\: {y}^{2} = 16x:⟼y2=16x</p><p></p><p>We know,</p><p></p><p>If m is the slope of tangent to the Parabola is y² = 4ax, then equation of tangent is</p><p></p><p>\rm :\longmapsto\:y = mx + \dfrac{a}{m}:⟼y=mx+ma</p><p></p><p>If we compare the given Equation of Parabola y² = 16x with y² = 4ax, we get a = 4.</p><p></p><p>So,</p><p></p><p>Equation of tangent having slope 'm' to the parabola y² = 16x is given by</p><p></p><p>\rm :\longmapsto\:y = mx + \dfrac{4}{m}:⟼y=mx+m4</p><p></p><p>\rm :\longmapsto\:y = \dfrac{ {xm}^{2} + 4}{m} - - (1):⟼y=mxm2+4−−(1)</p><p></p><p>Now, for this line to be tangent to xy = - 4, we have</p><p></p><p>\rm :\longmapsto\: \dfrac{ {xm}^{2} + 4}{m} \times x = - 4:⟼mxm2+4×x=−4</p><p></p><p>\rm :\longmapsto\: {x}^{2} {m}^{2} + 4x = - 4m:⟼x2m2+4x=−4m</p><p></p><p>\rm :\longmapsto\: {m}^{2} {x}^{2} + 4x + 4m = 0:⟼m2x2+4x+4m=0</p><p></p><p>For this line to be tangent, Discriminant = 0</p><p></p><p>i.e. b² - 4ac = 0</p><p></p><p>Here,</p><p></p><p>a = m²</p><p></p><p>b = 4</p><p></p><p>c = 4m</p><p></p><p>So, on substituting the values, we get</p><p></p><p>\rm :\longmapsto\: {4}^{2} - 4 \times 4m \times {m}^{2} = 0:⟼42−4×4m×m2=0</p><p></p><p>\rm :\longmapsto\:16 - 16 {m}^{3} = 0:⟼16−16m3=0</p><p></p><p>\rm :\longmapsto\:16 {m}^{3} = 16:⟼16m3=16</p><p></p><p>\rm :\longmapsto\: {m}^{3} = 1:⟼m3=1</p><p></p><p>\bf\implies \:m = 1⟹m=1</p><p></p><p>On substituting the value of 'm', in equation (1), we get</p><p></p><p>\rm :\longmapsto\:y = \dfrac{ {x}(1)^{2} + 4}{1}:⟼y=1x(1)2+4</p><p></p><p>\rm :\longmapsto\:y = \dfrac{ {x} + 4}{1}:⟼y=1x+4</p><p></p><p>\rm :\longmapsto\:y = x + 4:⟼y=x+4</p><p></p><p>Hence,</p><p></p><p>The equation of common tangent to</p><p></p><p>\rm :\longmapsto\: {y}^{2} = 16x:⟼y2=16x</p><p></p><p>and</p><p></p><p>\rm :\longmapsto\:xy = - 4:⟼xy=−4</p><p></p><p>is</p><p></p><p>\bf :\longmapsto\:y = x + 4:⟼y=x+4</p><p></p><p>$

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