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The parabola \( y=4-x^{2} \) has vertex \( P \). It intersects \( x \)-axis at \( A \) and \( B \). If the parabola is translated from its initial position to a new position by moving its vertex along the line \( y=x+4 \), so that it intersects \( x \)-axis at \( B \) and \( C \), then abscissa of \( C \) will be : (a) 3 (b) 4 (c) 6 (d) 8 |
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Answer» y = 4- x2 intersects x- axis at A and B ∵ 4 - x2 = 0 gives x = -2 or 2 Hence, (-2 ,0) & (2 ,0) are coordinates of A& B Replace x by x + y in equation (i), we get y = 4- (x+ 4)2 ⇒ 4 - (4 + 4)2 = 0 gives intersection points with x- axis ⇒ (x+4)2 4 = (+/- 2)2 ⇒ x + 4 = -2 or x+4 = 2 ⇒ x = -6 or x = -2 Hence, (-2,0) & (-6,0) are coordinates of B & C ∴ B = (-20) , A (2,0) C & (-6,0) ∴ abscissa of c will be -6 |
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