Find the equation of tangents to the curve `4x^2-9y^2=1`
which are parallel to `4y=5x+7.`
Text Solution
Verified by Experts
Let m be the slope of the tangent of `4x^(2)-9y^(2)=1.` Then `m=("Slope of the line "4y=5x+7)=(5)/(4)` We have `(x^(2))/(1//4)-(y^(2))/(1//9)=1` where `a^(2)=(1)/(4),b^(2)=(1)/(9)` The equations of the tangents are `y=mx pm sqrt(a^(2)m^(2)-b^(2))` `"or "y=(5)/(4)xpmsqrt((25)/(64)-(1)/(9))` `"or "30x-24y pm sqrt(161)=0` `"or "24y-30x=pmsqrt(161)`
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