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  • Can someone explain - Limit , continuity and differentiability - JEE Main-13

Equation of tangent at (2, 4) being on y^{2} = 8x is

  • Option 1)

    x - y + 2 = 0

  • Option 2)

    x + y -6 = 0

  • Option 3)

    2x + y = 8

  • Option 4)

    2x = y

 

Answers (1)

best_answer

As we have learnt,

 

Equation of tangent -

Let  (x_{\circ }\:y_{\circ })  is the point on the curve y = f(x) then direct we can write as 

x^{2}<xx_{\circ },\:\:\:\:\: x=\frac{x+x_{\circ }}{2}


y^{2}<yy_{\circ },\:\:\:\:\: y=\frac{y+y_{\circ }}{2}

for\:\:\:x^{2}+y^{2}=2,\;\:equation\:of\:tangent\:at\:(1,\:1)\:is\:x\times 1+y\times 1=2

\therefore \:x+y=2

-

 

 Equation of tangent at (x_1, y_1) will be:

yy_1 = 8 \left(\frac{x+x_1}{2} \right )

\Rightarrow y(4)= 4 (x+ 2) \Rightarrow x-y + 2=0

 


Option 1)

x - y + 2 = 0

Option 2)

x + y -6 = 0

Option 3)

2x + y = 8

Option 4)

2x = y

Posted by

Himanshu

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