If tangent to the curve y = log x at (e², λ) and normal to the parabola y² = 8x at (2, 4) cuts the y axis at the same point on T. Then λ is ______.

2 Views | Posted 3 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

3 weeks ago

Equation of tangent of at $y = l o g ? x$ at $(e^{λ}, λ)$ is $y - λ = \frac{1}{e^{λ}} (x - e^{λ})$

It cuts y axis at $(0, λ, - 1)$

Also normal to $y^{2} = 8 x$ at $(7,4)$ is $y - 4 = - (x - 2)$

$y = - x + 6$ Cuts y axis at $(0,6)$ SO ATQ $6 = - 1 + λ \Rightarrow λ = 7$

Equation of tangent of at <math> <mi>y</mi> <mo>=</mo> <mi mathvariant="normal">l</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">g</mi> <mo>?</mo> <mi>x</mi> </math>   at <math> <mfenced separators="|"> <mrow> <mrow> <msup> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> </msup> <mo>,</mo> <mi>λ</mi> </mrow> </mrow> </mfenced> </math>   is <math> <mi>y</mi> <mo>-</mo> <mi>λ</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mfenced separators="|"> <mrow> <mrow> <mi>x</mi> <mo>-</mo> <msup> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> </math>  It cuts y   axis at <math> <mo>(</mo> <mn>0</mn> <mo>,</mo> <mi>λ</mi> <mo>,</mo> <mo>-</mo> <mn>1</mn> <mo>)</mo> </math>  Also normal to <math> <msup> <mrow> <mrow> <mi>y</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>=</mo> <mn>8</mn> <mi>x</mi> </math>    at <math> <mo>(</mo> <mn>7,4</mn> <mo>)</mo> </math>    is <math> <mi>y</mi> <mo>-</mo> <mn>4</mn> <mo>=</mo> <mo>-</mo> <mo>(</mo> <mi>x</mi> <mo>-</mo> <mn>2</mn> <mo>)</mo> </math>     <math> <mi>y</mi> <mo>=</mo> <mo>-</mo> <mi>x</mi> <mo>+</mo> <mn>6</mn> <mrow> <mi mathvariant="normal"> </mi> </mrow> </math>  Cuts y axis at <math> <mo>(</mo> <mn>0,6</mn> <mo>)</mo> <mrow> <mi mathvariant="normal"> </mi> </mrow> </math>   SO ATQ <math> <mn>6</mn> <mo>=</mo> <mo>-</mo> <mn>1</mn> <mo>+</mo> <mi>λ</mi> <mo>⇒</mo> <mi>λ</mi> <mo>=</mo> <mn>7</mn> </math>