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The slope of the line touching both the parabolas `y^2=4x and x^2=−32y` isA. `1//2`B. `3//2`C. `1//8`D. `2//3` |
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Answer» Correct Answer - A 1 Equation of tangent to `y^(2)=4x` at A `(t^(2),2t)` is `yt=x+t^(2)` This is tangent to `x^(2)+32=0` `rArrx^(2)+32((x)/(t)+t)=0rArrx^(2)+(32)/(t)x+32t=0` Above equation must have eqaution roots `rArr((32)/(t))^(2)-4(32t)=0` `rArr32((32)/(t^(2))-4t)=0` `rArrt^(3)=8rArrt=2` `rArr" Slope of tangent is "(1)/(t)=(1)/(2)`. |
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