91Ó°ÊÓ

First, convert the given polar curve into rectangular coordinates using the conversions \(x=r\cos \theta\) and \(y=r\sin \theta\). The equation in polar coordinates is \(r=a-\cos\theta\). After substitution, we get: \(x=(a-\cos\theta)\cos\theta\) \(y=(a-\cos\theta)\sin\theta\) Now, we need to find the points where the curve intersects the pole. This will happen when \(r=0\), which means \(a-\cos\theta=0\). Solving for \(\theta\), we get: \(\theta=\pm\cos^{-1}(a)\)

Now, we need to find the slope of this curve. We can find this by differentiating both \(x\) and \(y\) with respect to \(\theta\) and then dividing \(\frac{dy}{d\theta}\) by \(\frac{dx}{d\theta}\). Differentiating \(x\) with respect to \(\theta\), we get: \(\frac{dx}{d\theta}=\frac{d}{d\theta}((a-\cos\theta)\cos\theta)=-\sin\theta\cos\theta-(a-\cos\theta)(-\sin\theta)\) Similarly, differentiating \(y\) with respect to \(\theta\), we get: \(\frac{dy}{d\theta}=\frac{d}{d\theta}((a-\cos\theta)\sin\theta)=\sin^2\theta-(a-\cos\theta)(\cos\theta)\) Now, we find the slope of the curve by dividing \(\frac{dy}{d\theta}\) by \(\frac{dx}{d\theta}\): \(m(\theta) = \frac{\sin^2\theta-(a-\cos\theta)\cos\theta}{-\sin\theta\cos\theta-(a-\cos\theta)(-\sin\theta)}\)

To find the right-angled intersection at the pole, we need to find the value of \(a\) for which the slopes are perpendicular when the curve intersects the pole. The slopes are perpendicular when their product is equal to -1, i.e., \(m(\cos^{-1}(a)) \cdot m(-\cos^{-1}(a))=-1\). Let's plug the intersection points obtained in Step 1 and evaluate the product of the slopes at those points: \(m(\cos^{-1}(a))\cdot m(-\cos^{-1}(a)) = \frac{(\sin^2\cos^{-1}(a)-(a-\cos\cos^{-1}(a))\cos\cos^{-1}(a))(-\sin\cos^{-1}(a)\cos\cos^{-1}(a)-(a-\cos\cos^{-1}(a))(-\sin\cos^{-1}(a)))}{(\sin^2(-\cos^{-1}(a))-(a-\cos(-\cos^{-1}(a)))\cos(-\cos^{-1}(a)))(-\sin(-\cos^{-1}(a))\cos(-\cos^{-1}(a))-(a-\cos(-\cos^{-1}(a))(-\sin(-\cos^{-1}(a))))}\) Now, let's simplify it: \((-1-\sqrt{1-a^2})(1-\sqrt{1-a^2})=-1\) Expanding and simplifying this equation, we obtain: \((-1+a^2)+2\sqrt{1-a^2}=1\) Dividing both sides by 2 and solving for the square root term, we get: \(\sqrt{1-a^2}=a\) Squaring both sides and rearranging, we get: \(a^4-2a^2+1=0\) This is a quadratic equation in \(a^2\): \((a^2-1)^2=0\) From this, we find that \(a^2=1\), which gives us two solutions: \(a=1\) and \(a=-1\). However, we are given that \(0<a<1\), so the only suitable value for \(a\) is \(a=\boxed{1}\).