Show that the curves 6x^2-5x+2y=0, 4x^2+8y^2=3 touch each other at (1/2,1/2).

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If two curves touch at 12,12, then slope of tangent at point 12,12 will be same for both the curvesWe have:6x2-5x+2y=0Differentiating with respect to x, we get:12x-5+2.dydx=0dydx=5-12x2dydx12,12=5-12×122=5-62=-12=m1We have4x2+8y2=3Differentiating with respect to x, we get:8x+16y.dydx=0dydx=-8x16y=-x2ydydx12,12=-122.12=-12=m2As m1=m2=-12Hence both curves touch each other at 12,12.

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